Welcome to Caltrans LSIT LS exam preparation course. One aid in your preparation for California licensure examinations. A word of caution, don't use this course as your only preparation. Devise and follow a regular schedule of study which begins months before the test. Work many problems in each area, not just those in this course's workbook, but problems from other sources as well. This course is funded by Caltrans, but you and I owe a profound thanks to others, the courses instructors, from the academic community, the private sector, other public agencies, and from Caltrans as well. We wish you well in your study toward becoming a member of California's professional land surveying community. Welcome viewers to this unit of the Caltrans training videos. My name is Edward Zimmerman. I work in the Geometronics branch at Caltrans headquarters in Sacramento. In the past, I worked for the California State Lands Commission, a private surveying firm in the city of Sacramento. I also teach surveying at Sacramento City College. I am a licensed as a professional land surveyor in California and in the state of Arizona. The subject I would like to discuss today is understanding and most importantly, simplifying and working with basic surveying mathematics. I want to emphasize that this video unit's objective is to present basic mathematical skills useful to surveyors. The discussion will be limited and is not intended as a comprehensive presentation of all surveying math procedures. It is taken for granted that most viewers are already familiar with some or most of the topics presented in the beginning of this unit. An understanding of basic arithmetic, algebra, geometry and trigonometry calculation methods is necessary to understand the more complex forms of mathematics that you will eventually encounter in your surveying career. Hopefully, this unit will help the surveyor and the trainee to develop and hone their computational skills and to make correct and effective use of appropriate solutions to particular survey calculation problems. I am assuming that the viewer is comfortable using and performing basic arithmetical operations. A review of the technical math test texts in the reference section of this workbook will provide additional help with operations and fundamental mathematical skills not covered in this video. In this day and age of electronic marvels, every surveyor has one or several types of computers and calculators. This video will present various principles of math to give the viewer a familiarization and a better understanding of the principles involved in solutions. It is assumed that a calculator will be used to determine numerical values of any given problems. The development of the decimal system can be tracked back many centuries. In ancient times, a small group of objects were counted by comparing their number with a number of fingers on one's hand. When larger groups of objects were counted, pebbles were arranged in groups of 10 to keep track of the large number of objects being counted. The modern counting system or decimal system is based on these early beginnings. The whole number part of a number falls on the left of the decimal point and the fractional part of a number falls on the right of the decimal point. The chart in this graphics shows the value of numbers both left and right of the decimal point. As shown, numbers to the left are whole and follow powers of 10 while numbers to the right are fractional. When adding a group of decimal numbers, tabulate them so the decimal point of each number is in vertical alignment up and down the column of numbers. For example, add 102.21, 14, 15.5, 21.348, 608.17, 6.411. Starting with a column to the right, total each column placing the decimal point of the total in line with the decimal point above. When subtracting decimal numbers, also tabulate the numbers so each decimal point is aligned vertically. For example, subtract 1324.465 from 21647.9. Starting with the column to the right, subtract each number and then place the decimal point of the difference in line with the decimal points above it. Notice that zeros have been added to the menu end to fill out the column. To multiply decimal numbers, determine the number of decimal places in both the multiplier and the multiplicand. The product will have the same number of decimal places as total places in the two factors. For example, multiply 843.21 by 56.893. The product will be 48,048.63543. In the second example, 0.043 is multiplied by 0.0021. Note that zeros must be placed between the product and the decimal point to create the correct number of decimal places. Long division of decimal numbers is set up and tabulated the same as whole division, whole number division. For example, divide 2467.89 by 10.45. Move the divisor's decimal point to the right enough places to make it a whole number. Move the decimal place in the dividend the same number of places to the right. Position the decimal place in the quotient directly above the decimal point in the dividend. You will generally have to add zeros to the dividend to attain the required numbers of decimal places in your answer. In the surveying measurement process, it is important to indicate the measurement, accuracy of a measurement by recording the correct number of significant figures. The number of significant figures in a quantity are defined as those digits that are certain or directly read plus one digit that is estimated and or uncertain. Significant figures are sometimes confused with decimal places. Decimal places do not indicate significant figures but are used to maintain the correct number of significant figures. Some examples are being shown in this graphic. In surveying calculations, it is important that calculations be consistent with the measured values. In addition and subtraction operations, answers should be rounded off to show the same digit as the digit expressing the least significant figure in the column above as shown in this graphic. As shown in this next graphic and multiplication or division operations, the least number of significant figures in the answer should not exceed the least significant figures in any of the factors. Rounding off numbers is the process whereby one or more digits is dropped from a number so that it contains only the required number of significant figures. When rounding off numbers to a required degree of accuracy, one of the following methods shown on screen may be used. First, when the digit to be dropped is less than five, the number remaining is simply stated without the digit. Second, when the digit to be dropped is exactly five, the nearest even number is used for the preceding digit. This method establishes a consistent pattern and yields a better balance of results. Last, when the digit to be dropped is more than five, the nearest preceding digit is increased by one. Simply stated, percentage is an indication of a number of hundreds to a whole. If a square contains 100 smaller square parts, then each of the smaller squares are one one hundredth of the whole or one percent. A simple percentage problem has three parts, rate, base, and percentage. In the graphic, rate is the percent figure, the base is the total quantity, and the percentage is the quantity of the percent of the whole. Thus, when computed, 10% of 120 is 12. Note that the decimal point in the percent is moved two decimal places to the left to yield the correct number for the percentage. There are three types of percentage problems. We have already seen how to determine the percentage when given the base and rate. Now let's look at the next one, finding the percent given the base and percentage. In this graphic, let's determine what rate or percent of 840 is 21. Just divide 21 by 840, which equals 0.025, or when the decimal point is moved two places right, 2.5%. In this example, the last of our percentage problems, given the rate and percentage, we will find the base or whole. As done in the graphic, change the rate by moving its decimal point two places left and divide the percentage by it. 915 is 75% of what number? The answer being 1220. A number squared is simply a number multiplied by itself, such as 8 times 8 equals 64. 64 being the expression of 8 squared, or the base number raised to the second power. A number that is cubed has been multiplied twice by itself, such as 12 cubed is 12 times 12 times 12 equals 1,728. 1,728 is the expression of 12 to the third, or 12 to the third power. Power is also called an exponent and signifies how many times a base number is to be multiplied by itself. If an exponent is five, this signifies a number raised to the fifth power. As shown in the graphic, 21 to the fifth equals 4,084,101. All handheld calculators perform exponential operations with simple keystrokes, so we will spend no further time on the topic. Surveyors generally measure linear and vertical dimensions in decimal survey feet, angular dimensions in degrees, and units of area in acres and square feet. However, it is often necessary to convert these measurements to other measurement units, such as decimal feet to feet and inches. I will next discuss conversion operations that will make conversions simple and accurate. First, let's consider the US survey foot and its relationship to the international meter and foot. The official unit of length in the United States is the modern international meter defined at a world conference in 1959. This created a new definition for the foot in the United States called the standard foot. All surveying prior to 1959 had been done on the old foot, so it was decided to retain it and rename it the US survey foot. The graphic now on screen shows the correct relationship of US survey foot and the new international meter. Any conversion operation regarding meters and survey feet should carefully consider the correct constants. Surveyors normally work in feet and decimals. However, it is sometimes necessary to convert dimensions to feet and inches and vice versa. As shown in this graphic, convert 123 feet, seven and three-quarter inches to decimal feet. In the first step, reduce seven and three-quarter inches to decimal inches by reducing three-quarters of an inch to its decimal form, 0.75 inches. In the second step, 7.75 inches is reduced to a decimal foot by dividing it by 12, adding the decimal foot to 123 equals 123.65 feet. The reverse of foot and inch to decimal, or decimal to feet and inches, is shown in this graphic. Convert 46.37 feet to feet and inches. First take 0.37 feet and multiply it by 12, converting it to 4.44 inches. This decimal fraction reduces to four and 44 100ths of an inch, or most nearly three-eighths of an inch. Adding the 46 feet back into the conversion completes the operation at 46 feet, four and three-eighths of an inch. The number of decimal places rounded to will be dictated by the accuracy required for the work. During colonial times, the most popular measuring device in the United States was the Gunters chain. It was 66 feet long and was divided into 100 lengths, each 7.92 inches in length. As can be seen in this graphic, it was very well suited for surveying work in English units. Converting feet to chains is accomplished by dividing the number of feet by 66. On the graphic, this operation yields 26.09 chains or 26 chains, nine lengths. In this example, we are converting chains to feet. Simply done, multiply the distance expressed in chains and lengths by 66 feet. The previous graphic showed the relationship of area to chain units, or 10 square chains equals one acre, or 43,560 square feet. The bottom conversion problem on the present screen converts a rectangular parcel measured in chains to area. The sexagismal system, or base-60 system, of angular measurement is used by surveyors in the United States and in many other parts of the world. This system is based on a complete circle being divided into 360 parts or degrees. Each degree is subdivided into 60 parts or minutes, with each minute divided into 60 seconds, as shown in the graphic. The European system divides the circle into 400 parts called grads. Each grad is subdivided into 100 parts or centigrades. Centigrades are further divided into 100 parts called melligrades. When using degrees, minutes, and seconds in calculations, it is often necessary to convert them to decimal form and vice versa in order to make the numbers compatible with the calculators being used. For instance, as shown here, 342135 converts to 34.35972 degrees in decimal format, or 96.39167 degrees converts to 96 degrees, 23 minutes, 30 seconds. Conversions of degrees to decimal form is also shown in this graphic. Conversion from decimal format, 63.31784 degrees to degrees, 63.31784 degrees to degrees, 63.1904 seconds is shown in this graphic. Remember, it is always required to convert to decimal form before using most calculators, unless they are pre-programmed to make the degrees to decimals and vice versa conversion. Radian measure is another definition of angular measure and has proven quite convenient for extensive calculations such as those involving circular curves and statistical analysis of sets of angles. The basic relationship of radians to degrees is shown in this graphic. Remember to convert degrees to decimal form when converting to radians. The most commonly accepted unit of area in land surveying is the acre which contains 43,560 square feet. Other units of area and their relationship to acres in square feet is shown in this graphic. Conversions to corresponding units may be made by inspection of this table in your workbook. This portion of the unit will center on random error. I must caution the viewer that time prevents a complete discussion of all types of survey measurement error other than random. I recommend additional study outside the training video to completely understand error in survey measurement. A random measurement error is defined as the difference between a measured quantity and its true dimension. Human limitations, limitations on the manufacturing position of survey instruments and other uncertainties act in concert to prevent ever making a true measurement. Therefore, it can be stated that no measurement is exact and will always contain some error of unknown size. It is impossible to prevent error, and it should be an objective of the surveyor to recognize error and control it. Random errors are unavoidable. They follow laws of probability and tend to be of either sign and are quite small. Hence, they tend to almost but not quite cancel each other out. Making a set of measurements of the same quantity and then studying the results will indicate differences in value and express the reliability of each measurement by showing its difference from the mean of the set. An analysis will determine and quantify the mean or most probable value of the set to truth. However, before we explore this process, let's define some of the terms involved. In the first term shown on the graphic is most probable value. The most probable value is also the arithmetical mean or average of the set. The next term considered is shown in this graphic. It is called the residual and is the difference between a single measurement and the most probable value. Residuals play a key role in the analysis process that we will be looking at in a minute. This graphic shows the next term to be defined, the standard error. This factor defines the value or accuracy of a set of measurements and sets the limit within which 68% of the measurements will fall. The probable error modifies the standard error and is shown on this screen. It fixes the limits between which 50% of the measurements within a set would fall. Other modifications are shown and are known as the 90% error and the 95% error, which set limits within which 90% and 95% of the measurements respectively will fall. The most probable value of a set of measurements is the mean or average. As you will see, the mean is derived from the set, so it, too, is subject to random error. The mean is often accepted as a final result, so its dependability should be quantified. The equation for standard error of the mean is shown on this graphic. The standard error of the mean can be modified to reflect the 90% error and the 95% errors, also by applying the same modifiers shown in the last graphic. To put all of our foregoing discussion of equations and error computation into use, let's take a look at the current graphic. Totalling the lengths equals the summation of 6,777.42 feet. This divided by 6 and number of measurements yields the most probable value of 1,129.57 feet. Determining the residuals, squaring them and then totalling them equals a residual summation of 0.0168. When this figure is plugged into the equation for standard error of the set, the result is plus or minus 0.05 feet rounded to two places. This means that 68% of the measurements within the set will fall between 1,129.52 feet and 1,129.62 feet. The standard error of the mean was found to be plus or minus 0.02 feet, which signifies there is a 68% probability that the true distance lies somewhere in the range between 1,129.55 feet and 1,129.59 feet. Until this point, we have been discussing errors affecting only a single measurement. Random error also affects measurements resulting from combining two or more quantities by computational processes. A sum or product of separate measurements, each with its distinct random error, will have a total random error influenced by a composite of the individual errors. The first propagation of error process that we will look at is identifying the error created in a sum of a number of several measurements made under different independent conditions. A line has been measured in four segments, each segment under different conditions yielding a different standard error for each segment. In this graphic, the estimated standard error for the sum of the measurements is determined to be plus or minus 0.054 feet. Next to be considered is the error developed in multiplication of measurements such as for determining areas. This graphic illustrates the determination of expected error in a calculated product. The completed calculation states the product has a standard error of plus or minus 3.02 square feet. This error occurs when a series of measurements made under the same conditions are totaled up. Each of the measurements are subject to the same random error and tend to cancel each other out. The equation for this computation is shown in this graphic. This process states the distance 2,175 feet was measured with an accuracy of plus or minus 0.12 feet. I would remind the viewer again that random error is inevitable in survey measurement. Live with it and learn to identify and control it. Do not waste time trying to eliminate it. Algebra can be considered an extension of arithmetic in which symbols represent unknown quantities related to known factors of an expression. All knowns and unknowns are integral to operations that are consistent to numbers in the expression. I include the following discussion to develop a basic awareness in solving for unknowns and to develop the worker's skill in working with negative numbers. Numbers in mathematics are of three categories. Positive numbers that are greater than zero, negative numbers that are less than zero, and zero itself, which needs no explanation. As shown in the sketch, signed numbers to the left of zero are considered negative and preceded by a minus sign. Numbers to the right of zero are positive numbers and may or may not be preceded by a positive sign. The absolute value of a number is its value without reference to a sign. Thereby, positive 6 would have the same absolute value as negative 6. In working with signed numbers, there are certain guidelines to follow in avoiding blunders and getting answers with incorrect signs. First, let's consider the addition of signed numbers. As shown in this screen, the sum of two positive numbers is positive. 46 plus 34 equals 80. The sum of two negative numbers is negative. Minus 21 plus minus 76 equals minus 97. The sum of a positive and negative number will be first determined the difference in absolute value of the number. Positive if the absolute value of the positive number is larger. 82 plus negative 31 equals 51. Negative if the absolute value of the negative number is larger. 26 plus negative 67 equals negative 41. The next case of signed number operations, the subtraction operation, the rule is rather uncomplicated when subtracting one signed number from another, simply reverse the sign of the second number and then add the two numbers using the addition rules just discussed. Multiplication of signed numbers is also a very simple rule to follow. The product of two positive numbers is always positive, while the product of a negative number and a positive number will always be negative. However, the product of two negative numbers will always be a positive number. The last case of operations with signed numbers concerns division. The rule for division is the same as multiplication. The quotient of two positive numbers is positive, while the quotient of two negative numbers is also positive. The quotient of a negative number and a positive number is always negative. An algebraic equation should be thought of as a math statement that is balanced on both sides of its equal signs, or quantities on the left side must equal the quantities on the right side. Each side of an equation may be modified, but remember to modify each side by exactly the same degree and quantity, or do the same thing to both sides. An equation is used to find the value of an unknown number from its relationship to the known values in the equation. The ability to visualize a mathematical problem in words and then form it into an equation is a valuable skill for the surveyor. For example, in this graphic, unknowns are identified as B or Z, and then equations are formed from the word statements to reach solutions. As shown by the two examples in this graphic, equations are formed from conditions set by a word statement. A number divided by three plus the same number times seven equates to B divided by three plus B times seven. A number plus 212 times eight plus the same number divided by 34 equates to Z plus 212 times eight plus Z divided by 34. When working with more than two quantities in an algebraic expression, perform the indicated work in the order presented in the next graphic. Work from the left doing multiplication and division. Work from the left again doing addition and subtraction operations. For example, 267 times six plus 96 divided by four minus four times 12 minus nine reduces to 1,602 plus 26 minus 48 minus nine equals 1,571. The second example, 301 minus 22 times 14 plus 81 divided by 27 minus 76 reduces to 301 minus 308 plus three minus 76 equals negative 80. In algebra, it is often necessary to group terms and consider a whole group of several groups as a single term. Grouping is indicated by parentheses, brackets, or braces. The plus or minus sign in front of each group applies to each term in the group. Rules for removing parentheses are presented in this graphic. If parentheses are preceded by a positive sign, they can be removed without changing the signs of terms inside. If preceded by a negative sign, remove the parentheses by changing the sign of every term within the group. It is much like multiplying a group by plus one or by negative one. First, 5x plus 3y minus 4z equals 5x plus 3y minus 4z. Second, 5x minus 3y minus 4z is reduced to 5x minus 3y plus 4z. Oftentimes there are several groups in one problem. In this case, remove the intercept first and then the outer set and so on following the prior stated rules. An algebraic equation contains signs for some of all the four mathematical operations. Add, subtract, multiply, or divide. And also letters that are substitutes for numbers. These letters are called variables and when replaced by actual number values through substitution, the equation is simplified. When simplifying an equation, simply plug assigned numbers into the variables and then combine the terms following the rules of order of operations. As shown in this graphic, make the following substitutions. Let A equal 12, B equal 6, and X equal 7. X plus A times B minus B equals 7 plus 12 times 6 minus 6, or 7 plus 72 minus 6 equals 73. Solving equations. Equations are generally solved by picking out one of the unknowns, isolating it, and then solving for it. Something to keep in mind, it is much easier to manipulate the variables and reach a formulated solution prior to substituting actual numerical values. For example, let's solve this equation for R, assuming we already know the value of A. Let's follow the solution on the graphic, keeping in mind to do the same thing to both sides of the equation. Area of a circle is A equals pi times radius squared solved for radius. Let A equal 9,503 square feet. Rearrange R squared equals area divided by pi. Extract the square root, R equals the square root of area divided by pi. Up to this point, equations in the processes discussed have been elementary first power equations. In many problems, cube and higher powers will occur. A quadratic equation is a function which contains the second power of one unknown. A quadratic equation useful for surveying computations is generally written as AX squared plus BX plus C equals 0. A, B, and C are generally numbers with X remaining as a value sought. For ease of solution, the quadratic equation is reduced to the quadratic formula shown in this graphic. There is a plus or minus sign in front of the square root sign. This will yield two different answers when solving a quadratic. You must determine by a careful analysis of the problem which is the correct answer. Familiarity with algebra gives the surveyor the ability and tools to reduce a problem to a logical solution by arriving at and correctly setting up an algebraic formula. It is much easier to work with equations before they have the numbers plugged in. When the solution has been found in the equation set up, then and only then should the numbers come into play. I highly recommend keeping your calculations as simple as possible. Don't add any unnecessary steps. An extra step just increases your chances of making any blunder or calculation error. Geometry is as old as civilization. We know the Sumerians and the Egyptians had the knowledge of and applied the principles of geometry. In fact, a translation of the Greek word geometry means earth measurement. Geometrical facts, figures, and formulas are essential to understanding and applying mathematical processes to survey. The following definitions will provide a brief review of geometric principles. We will start our discussion with some of the basic corollaries or laws of geometry. Let's first talk about lines. As shown in example number one of this graphic, straight lines extend to infinity in both directions. Example number two shows that if straight lines in the same plane meet, they are intersecting lines and share a common point. Example number three proves that if two straight lines have no common point, they are parallel lines. A line extending to infinity from a defined starting point is a half line as shown in figure number four. As shown in figure number five, a line starting from a defined starting point and ending at another defined point is a line segment. Angles are formed when two half lines or segments are drawn from the same starting point. You can also form an angle by rotating a half line about its starting point, which is also the vertex of the angle formed. The dimension of the angle formed is measured and stated in degrees or radians. This graphic shows the different types of angles in their definition. As shown, angles are classified as acute less than 90 degrees, right exactly 90 degrees, obtuse between 90 and 180 degrees, straight exactly 180 degrees, reflex between 180 and 360 degrees. A pair of angles, the total 90 degrees are called complementary angles. This means that 37 degrees would be the complement to 53 degrees. Two angles that total up to 180 degrees are defined as supplementary angles. In other words, 68 degrees would be the supplement to 112 degrees. In geometry, certain basic theorems or statements exist to serve as basic rules. I am presenting some of the rules in the next few graphics to give you some more basic geometrical concepts about lines and angles. Proof of the theorems presented has been omitted from my presentation. The rules are sufficient to stand on their own. Rule number one, when two lines intersect, angles formed opposite to each other are identical. Thus, angle one equals angle three and angle two equals angle four. Rule number two, when two parallel lines are intersected by a transverse line, the alternate interior angles formed are the same. As shown, angle one equals angle three and angle two equals angle four. Rule number three, when two parallel lines are intersected by a transverse line, the corresponding angles formed are the same. As shown, angle one equals angle five, angle two equals angle six, angle three equals angle seven, and angle four equals angle eight. Rule number four, two angles are the same or supplement each other if their corresponding sides are perpendicular. As shown, angle one equals angle two and angle two plus angle three equals 180 degrees. Almost all surveying operations and computations involve geometrical figures. Therefore, I will be presenting most of the basic ones encountered in our work along with a brief look at some of the properties and rules of the figures. Let's start with our old friend, the circle. This often used figure is the basis of horizontal curve computation and layout as explained in another of the video units. A circle can be defined as a closed curve in the same plane, every point of which is the same distance or radius from a point within the circle called the center of the circle or radius point. In this graphic, the basic portions of a circle are identified. If you traveled entirely around the circle starting at point A and returned to point A, you would have walked the circumference of the circle. If you went from point A along the straight line to point C, which is the diameter which bisects the circle. The line from the radius point to point B represents the radius, which is one half the diameter. Back to walking around the circle again. The distance from point B to point C describes an arc. Also, if a line is drawn from the center of a circle, line from rp to point f, perpendicular to a cord, it bisects the cord and the arc defined by the cord. This important rule comes into play when working with circular highway curves. Sector and segment are another two elements of a curve that should be familiar to a surveyor, as they are often used to calculate areas of surveys with curved boundaries. As shown in this graphic, the sector of an arc is the shaded wedge-shaped figure bounded by its arc and two radial lines. Its area is determined by the formula shown in the graphic. Sector area equals radius squared times delta expressed in radians divided by two. Note on the graphic that the radius of the arc is 139 feet and the delta angle is 47 degrees 30 minutes. Let's stop the tape at this point and work the problem. Okay, you have turned the tape back on. Let's check your answer. I get 8009 square feet. What did you get? This screen shows the second curve element under discussion, the segment of an arc. It is the shaded area and is defined as bounded by its arc and the cord for its arc. The area of a segment is calculated by the method now on screen. First, determine the area of the sector of an arc, OPR, by the formula just discussed. Sector area equals radius squared times delta expressed in radians divided by two. Then determine area of triangle OPR. The formula for this triangle is radius squared divided by two times sine delta. When both areas are computed, subtract the area of the inscribed triangle from the sector area, thereby finding the area of the segment. Letting R equal 421.25 feet and delta equaling 28 degrees 31 minutes and 15 seconds, shut the tape off and compute the area of segment OAC. Then turn the tape back on to check your answer with mine. Okay, I got 44,166 square feet for the sector area and 42,365 square feet for the inscribed triangle. Subtracting the triangle from the sector gave me the area of the sector 1,801 square feet. Is this what you got? Some other formulas dealing with curves are shown in this graph. They are circumference equals 2 times pi times radius. Area equals pi times radius squared. Radius equals 0.56419 times square root of area. Diameter equals 1.12838 times square root of area. Well, this pretty well finishes the discussion of circles, so let's move on and discuss some more geometrical figures that you will likely encounter in your surveying and test-taking activities. A polygon defines any figure bounded by three or more line segments. A polygon may have any number of sides as long as they are broken. That is segmented and enclose a figure. The simplest polygon is the triangle having the least number of sides, three. The triangle is one of the most used figures in surveying and survey calculation. It has many configurations in classification. Let's take a look at them. Triangles are classified in two different ways, by sides and by angles. First, we will look at the triangles classified by sides. The first of three types of triangles classified by sides are shown as figure one in this graphic. This triangle is called an isosceles triangle and is distinguished by having two equal length sides. Note that the unequal side is called the base and that the angle opposite the base is the vertex angle. The second type of triangle in this classification is shown in figure two and is called an equilateral triangle. This one has all three sides of equal length. The third and last triangle in this classification is the scaling triangle and it has three unequal sides as shown in figure three in this graphic. Triangles are also classified by the angles they contain. There are three types of triangles within this classification. Acute, obtuse, and right. We will look at figure one in this graphic and discuss the acute triangle first. This type of triangle has three acute less than 90 degree angles. Type number two, figure two, has one angle of the three that is an obtuse triangle, more than 90 degrees, less than 180 degrees. The other two angles of the triangle are always acute. The last type in the angular classification has one angle that is exactly 90 degrees. The other two angles of the right triangle will always be acute. As shown in figure three of this graphic, the side opposite of the right angle would be called the hypotenuse. The right triangle is the workhorse of survey measurement and calculation and will be discussed in greater detail later in this video unit. The next few graphics show polygons other than triangles. These are polygons having more than three sides. Familiarity with any of these polygons will prove useful to the surveyor and test taker. Before we look at any of the polygons, let me state a golden rule concerning interior angles of any polygon. The sum of the interior angles of any polygon is equal to 180 degrees times the total number of sides less two. This formula provides a check of interior angles measured for a closed figure. The first polygons to be considered are the ones defined as regular or having all sides and angles equal. As shown in these next graphics, polygons are named for their number of sides. A formula for area computation is also shown in each of the graphics. The first regular polygon to be considered is the equilateral triangle. All three sides the same length. Its area is easily computed by the formula area equals 0.25 times n number of sides times side squared times cotangent of 180 divided by n. Also shown is the symbology for the formula where we let n stand for the number of sides and s for the length of the sides which are identical in this case. The next regular polygon to be discussed in the series should have been the four-sided square. However, its properties are well known, so let's move on to the pentagon or five-sided figure. This graphic shows a regular pentagon, five equal sides and angles. Computation of its area is also by the same formula. In fact, this formula can be used to compute the area of any regular polygon regardless of the number of sides. Again, n equals number of sides and s equals the length of the sides which are identical in length. This graphic shows a regular hexagon, six equal sides and angles. Computation of its area is by the same formula. There is no need to look at any more regular polygons as the pattern of construction and area formula should be very clear by now. Surveyors should be interested in the additional polygons discussed in the next few graphics. They consist of various forms of quadrilaterals or four-sided figures. The square and rectangle will not be discussed as they should be very familiar to the viewer. The parallelogram is a quadrilateral with the opposing sides equal and parallel. All its angles are different than 90 degrees except in the case of a rectangle which has 90-degree angles. A parallelogram that is not a rectangle can also be considered a rhomboid. As shown in the graphic, the rhomboids area can be calculated by area equals a times b times sine theta. a and b symbolize the length of their respective sides. This graphic shows a parallelogram or rhomboid with all sides of equal length. This figure is a sketch of a rhombus. Its area is computed from the formula area equals b squared times sine theta letting b equal the length of the sides. A quadrilateral with only two of its sides parallel is named a trapezoid and is shown in this graphic. This configuration is one heavily used for engineering design and layout. Its area is determined from the following formula area equals h divided by 2 times b plus b prime. In this formula, b stands for the length of sides and h is the altitude. A trapezium is defined as a quadrilateral with none of its sides being parallel. Its area is calculated by the equation area equals d times h over 2 plus b over 4 times h prime plus h double prime where b equals the bottom width, d equals the top width, h the first altitude and h double prime the second altitude. The irregular pentagon is the last of the plane geometric figures I will discuss. This figure has five sides none of which are parallel or equal in length. Also, none of its angles are equal. Its area can be calculated by the equation on the graphic which is the same one discussed in the last graphic. I have reviewed many of the basic and fundamental facts found in geometry. Time prevents a complete review of a subject that takes many hours of study and practice to fully appreciate. However, the aspiring surveyor and or examinee should further pursue the study of geometry as I have only touched the surface of the subject. There are certain geometrical constructions that have a great value in solutions of surveying problems and calculations. Well, as much as I enjoy talking about surveying math and numbers, I feel that perhaps this is a good time to take a break. When we come back, we will launch into a study of trigonometry. Welcome back from break. If you're ready, we'll continue our discussion of survey mathematics with a look at trigonometry. Trigonometry is a natural extension of algebra and geometry and gives the surveyor the tools to conveniently solve problems involving unknown angles and distances. Trig enables the surveyor to take advantage of the relationship of angular dimension to distance measurement and put the principles of solving for indirect or missing measurements into use. The right triangle is the first and most important element of trigonometry that we will cover. It is the simplest of triangles and plays a major role in just about all the measurements and computations we do. A right triangle has been defined earlier as any triangle that includes an angle of exactly 90 degrees as one of its three angles. As shown in this graphic, the longest side of any right triangle is called the hypotenuse and lying opposite angle C is also called side C. The side opposite angle A, side A, is called the opposite side and the remaining side, which is opposite angle B, is side B or the adjacent side. The triangle is shown is the standard figure both in position and terminology for solution of all trigonometric functions and equations. The Pythagorean theorem applied only to right triangles states that the square of the hypotenuse of any right triangle is equal to the sum of the squares of the opposite side and adjacent side. As shown on the screen, the equation formed for this theorem is side A squared plus side B squared equals side C squared. This basic equation may be rearranged to provide solutions regardless of which side or the right triangle is sought. The arrangement has provided the other three equations which can be used to find whichever side is sought. As shown, side A is found by solving for the square root of side C squared minus side B squared. Side B is found by solving for the square root of side C squared minus side A squared. Side C is found by solving for the square root of side A squared plus side B squared. Let's try a practice solution for the problem shown in the upcoming graphic. Given the right triangle shown on screen solved for side C, at this point, pause the tape and solve the problem. Okay, now that you're finished, this is how I found the answer. First, rearranging the basic equation, we get C squared equals A squared plus B squared. Which reduces to finding the square root of the quantity 1,444 plus 7,396. It equals 8,840. The square root of 8,840 being 94.02 feet or the length of side C. Fairly routine, isn't it? In the right triangle, the difference in the lengths of two sides determines the size of their included angle. Therefore, the ratio between the lengths would determine the magnitude of certain trigonometric functions or ratios. The names and elements of the first three ratios of different sides of the right triangle to each other are shown in this next screen. Note that the ratios are defined in terms assigned to sides and angles of the standard right triangle we have already discussed and was shown a few graphics ago. Remember that the size of the triangles does not affect the magnitude of the functions. They are determined solely from the differences in two lines being compared.    Formulas for ratio or functions are made to be rearranged to satisfy specific needs or a required solution. Since functions are derived from sides of triangles the functions can be used to determine lengths of missing sides. All that is needed is one given side and a given angle to solve for a missing side. Missing sides of right triangles can be solved by the application of one of the rearranged equations shown in the next two graphics. Selection of which function to apply depends on which side or angle you want to solve for and what elements of the triangle are already known. As shown on this screen all the function ratios have been rearranged to provide a specific formula to solve for a particular problem. These are the first three to be discussed. Side A equals side C times sine of A or equals side B times tangent of A. Side B equals side C times cosine A or side A times cotangent A. Side C equals side A divided by sine A or side B divided by cosine A. This graphic shows the other three sets of rearranged trig functions for solving right triangles. Side A equals side C times cosine B or side B times cotangent B. Side B equals side C times sine B equals side A times tangent B. Side C equals side A over sine B equals side B over cosine B. Inspection of a problem and then a correct choice of function always yields the correct solution providing the simple math like multiplication and division correctly. There are two distinct operations when working with trig functions. One is to determine the numerical function for a given angle. The other is to determine an angle from a given numerical function. Electronic calculators provide a quick and accurate method for finding functions and arc functions. All calculators have keys marked sine, cosine, and tangent. Simply under the angle, generally in decimal of degrees format for most handheld calculators and press the key designated for the function sought. Set your calculator to read out to the nearest six decimal places in order to provide the correct accuracy to be consistent with today's surveying instruments. For example, in this graphic what is the value of the cosine function for 43 degrees, 6 minutes, 23 seconds? It is 0.730086. The second operation is what is the arc function value of a given numerical trig function? What is the arc sine value of 0.041893? It is 2 degrees, 24 minutes, and 6 seconds. Let's talk about finding missing elements of a triangle by using trig. For our presentation, the missing elements will fall into two categories. Solving for a missing angle with two sides known and solving for a missing side with one side and angle known. Remember that sine, cosine, and tangent values are unique to a particular bearing or angle. All the triangle equations can be manipulated to find any of the missing values as long as any two of the triangles of other values are known. Let's solve for a missing angle when using given data from a triangle. We want to solve for angle A given side B and side A as shown on this graphic. Side A equals 109.78 feet and side B equals 671.42 feet. Select an equation to fit the conditions. In this case, tangent A equals A divided by B. Please pause the tape at this point. Now plug the given quantities into the equation and perform the operation. Unpause the tape when you are finished and we will compare answers. Okay, you're back. Did you work the problem this way? 109.78 feet divided by 671.42 feet equals art tangent 0.163.505 which equals 9 degrees 17 minutes in 10 seconds. Is this what you got? The next example that we are going to work is solving for an unknown side. Application of the appropriate function from a given angle and side of our triangle and then computing the length of the missing side is the basic operation. Selection of the proper function to use is made easier if you make a quick sketch of the triangle showing given values and checking the relationship of the side sought to these given values. After choosing the best equation to use, determine the mathematically correct numerical function for the given angle and then perform the appropriate math operations with the given side selected to determine the missing value. Use six decimal places for all numerical trig functions that you use in the process. Round off your final answer to usually two decimal places. Let's try a missing side example. As will be shown in the upcoming graphic, we want to find side A. Information given is side B equals 312.47 feet and angle A equals 21 degrees 37 minutes in 5 seconds. On your sketch, note that only the given side is B and the only given angle is A. That means your equation will have to be one that contains these elements. For inspection of the equations, it appears that the best trace would be side A equals side B times tangent A. If you agree, pause the tape now and complete your solution and then come back on to check my solution.  Is this the same thing that you came up with? Let's do one more solution before leaving this topic. Given side C equals 636.02 feet and angle A equals 44 degrees 56 minutes and 35 seconds. Solve this triangle for side B. After looking this sketch over, I decided to leave the selection of a best equation to use for you. One more time if you want to, pause the tape and solve this example. When you have your answer, resume the tape and we can compare our notes. I use side B equals side C times cosine A. This reduces to 636.02 feet times 0.707809 equals 450.18 feet. This time, check your answer by the Pythagorean theorem by finding side A and then adding the squares of sides A and B for a check. I computed side A as 449.29 feet. This was the last of our right triangle problems. Not all triangles encountered in surveying are right triangles. So let's now move on and take a look at dealing with oblique triangles and put more of our trigonometric skills to work. As described earlier, an oblique triangle is one that has no right angles and no sides equal to each other. This makes the solution of obliques a bit more complex than the right triangle. The available solutions for oblique triangles can be grouped into four cases. Case number one, given two angles and a side opposite. Case number two, given two sides and the included angle. Case number three, given all three sides. Case number four, given two sides and the opposite angle. Generally, two formulas are used to solve oblique triangles. The law of signs and the law of cosines. The law of signs states that in any triangle, the sides are proportional to the sign of the opposite angle. The law of cosines states that in any triangle, the square of any side is equal to the sum of the squares of the other two sides minus twice the product of those two sides multiplied by the cosine of their included angle. The law of signs equation for oblique triangles is shown on the top portion of this graphic. As shown, side A over sine A is the same as side B over sine B equals side C over sine C. Equations based on the cosine law are also shown.  In the following discussion of solutions for oblique triangles, only basic equations will be addressed. The following will take some of the cases most likely to occur in surveying and briefly explore the equation for them. This graphic shows the configuration and notation for the oblique triangle solutions along with the first general equation we will be discussing. When given angle A and angle B along with side A, you can solve for the following.  The next solution occurs when angle A side B inside C are given, as shown in this graphic. The equation takes this form. To solve for A, A squared equals B squared plus C squared minus two BC times cosine A. The last general equation for oblique triangles is for solving for angle A when only the three sides of the triangle are given. In this case, no angles are given, but all three sides are. The equation to solve for angle A is cosine A equals quantity B squared plus C squared minus A squared over two BC. Remember that all elements of an oblique triangle can be rotated around the triangle to fit given conditions as long as they are kept in the same order to each other. Angles and directions are a basic element of survey. We should be familiar with the relationship of horizontal angles to the bearing and azimuth systems of angular definition. Bearings or azimuths, azimuths of survey lines must be referred to some north-south meridian in order to maintain consistency and adherence to horizontal reference system or horizontal data. Most horizontal systems are oriented to true north, which is a line on the surface of the earth that joins the north and south poles. However, some systems are referenced to magnetic north, a local assumed north, or a grid north. Regardless of the system's origin, bearings, azimuths, and horizontal angles all relate to each other in the same manner and proportion. For the purpose of discussion, all direct calculations in this presentation are assumed to be referred to true or geographic north. Let's first discuss the bearing system of direction. Bearings are a method of stating the direction of a survey line by indicating the angle of the line from its north-south meridian and also stating which quadrant of a line falls in. Bearings are always measured to the east or west from a north-south meridian line, and their values are always 90 degrees or less. Quadrant notation is indicated by N or S preceding the numerical value with E or W following the numbers. For instance, if the bearing of a line was measured 36 degrees, 45 minutes east of north, its direction is written as north, 36, 45 east. The north and east designation is just as vital as the numerical value. Four bearings are shown in this sketch. They each lie in different quadrants and are identified as follows. Line OA equals north, 36 degrees, 45 minutes east. Line OB equals south, 20 degrees, 14 minutes east. Line OC equals south, 46, 11 minutes west. Line OD equals north, 49 degrees, 37 minutes west. Note that measurement of bearings in the northeast quadrant starts at north and rotates in a clockwise direction to the east. While bearings in the northwest quadrant also start at north, they are measured in a counterclockwise rotation to the west. Bearings in the southeast quadrant start at south and are rotated in a counterclockwise direction to the east. Bearings in the southwest quadrant also start at south and are measured in a clockwise rotation to the west. Any survey line actually has two directions. Forward in the direction that a survey is progressing to have been. Normally, bearing designation follows a consistent direction of travel around a figure. For computational purposes it is often necessary to reverse the bearings. Easily done. As shown in this sketch, simply reverse the quadrant designators to turn the bearing around by 180 degrees. In this line, the line L to M has a forward bearing of 7104 east. Reversing the bearing, the same line has a back bearing of north 7104 west. When computing angles from adjacent bearings, the surveyor must routinely reverse bearings as computations proceed around in a longer figure. The azimuth system is another method of measuring the angular dimension of a survey line. Azimuth measurement originates at the north end of a meridian, 0 degrees, and rotates in a clockwise direction all the way around the circle back to north again, or 360 degrees. Azimuth values fall in a range from 0 degrees to 360 degrees. In this sketch, the line M to S is the meridian. Line O to A has an azimuth of 44 degrees. Line O to B has an azimuth of 135 degrees. Line O to C has an azimuth of 258 degrees. And line O to D has an azimuth of 335 degrees. As in the bearing system, every survey line has two azimuths, a forward azimuth and a back or reverse azimuth. When the azimuth of a line is designated, it is assumed that the angular value of the line is from the beginning of the line as one faces the end point of the line. Frequently for computational purposes, the surveyor must determine back azimuths of lines. Again, very simply done. Just add or subtract 180 degrees from the forward azimuth of a line. In the graphic now on screen, the forward azimuth of line D to E is 49 degrees 11 minutes based on the angle measured from the north and the direction of travel. Reversing the direction of travel adding 180 degrees determines the back azimuth of the same line as 229 degrees 11 minutes. In review, bearings are reversed by switching the letters of quadrants. Azimuths are reversed by adding or subtracting 180 degrees. Since they both originate at the same point, bearings and azimuths are easily interchanged. Azimuth directions are more convenient for computing directions and angles while most maps and survey tradition dictate the use of bearings. Therefore, a surveyor should be ready to make azimuth bearing conversions. Let's take a look at the next graphic while I discuss this process a little further. This graphic shows the angular relationship of azimuths to bearings in each of the four quadrants. For instance, look at the northeast quadrant. Line OA has a bearing of north 83 degrees 07 minutes east. Since azimuth and bearing both start at north and rotate clockwise, they will be identical in this quadrant. Next, we go to the southeast quadrant and take a look at line OB. Its bearing is south 3311 east. Remember, southeast bearings originate from south and rotate counterclockwise while azimuths rotate clockwise. An angle from north around the south is 180 degrees. Therefore, subtract the southeast bearing from 180 which equals 146 49 the azimuth of line OB. In the southwest quadrant, just add 180 degrees to the bearing value to obtain azimuth or subtract 180 degrees from the azimuth value to obtain the value of the southwest bearing. This should be routine by now, so most of you have already figured out that you subtract a northwest bearing from 360 degrees to obtain azimuth or subtract azimuth from 360 degrees to obtain the value of a northwest bearing. In review, bearing to azimuth conversion if it is a northeast bearing, it is the same as azimuth. Considering a southeast bearing subtract bearing from 180 degrees to find azimuth. Considering a southwest bearing add 180 degrees to bearing to get to azimuth. Considering a northwest bearing subtract bearing from 360 degrees to get azimuth. To do an azimuth bearing conversion, if it is in the northeast azimuth it is the same as the northeast bearing. If it is in the southeast quadrant subtract azimuth from 180 degrees to get a southeast bearing. If in a southwest quadrant add 180 degrees to bearing to get to azimuth. If it is in a northwest bearing subtract azimuth from 360 degrees to get northwest bearing. Most surveys require that surveyors compute angles from sets of given bearings or determine angles from horizontal angles or combining both operations in the course of computing. Let's take another and look at this next graphic and let's explore the process. Given a bearing and include an angle to a second survey line determine the bearing of the second line. At this point let me express how important a sketch is to the angle and bearing calculation process. It almost makes the problem self solving if you draw a sketch like the one on screen. Notice that all data is shown on the sketch along with an X for the missing value. In this case the bearing angle for line O to J from inspection of the sketch X equals the bearing angle of line OJ. Therefore X equals 103 degrees 31 minutes minus 58 degrees 39 minutes equals 44 degrees 52 minutes or north 44 degrees 52 minutes west. In this graphic the example presented and sketched is finding the included angle X between lines O to C north 31 36 west and O to A north 56 11 west. Look the problem over and then pause the tape solve the problem and then get back on the tape as we will compare answers. Ok this is what I did and this is the answer I got. X equals 180 degrees minus 56 degrees 11 minutes plus 31 degrees 36 minutes which equals 92 degrees 13 minutes. How did you do with this one? Like I said earlier one sketch is worth a lot of work. This graphic shows the last of the bearing angle examples and is an azimuth question. Note on the sketch that we are looking for X the included angle between the two azimuths. Once again pause the tape solve for X and get back to me with an answer. Welcome back. I found by using my sketch and the information on it that X equals 135 degrees 35 minutes. How did I get this answer? By subtracting 146 degrees 11 minutes from 281 degrees 46 minutes. You will be confronted with an almost limitless number of angle bearing azimuth problems on a daily basis. Any test you take is assuredly going to have this type problem in it. However, if you draw a sketch the correct position of all lines and all the knowns and unknowns any angle bearing azimuth problem can be almost self solving. Points located by surveys are referenced to a known point in one of two ways. By polar relocation or by rectangular location. Polar location ties a found point to a known reference location by a bearing and distance measurement as shown in this graphic. Note that point L is referenced to point K by a line of north 76 degrees 23 minutes east 611.21 feet. Rectangular location ties a found point to a known location by two distances that are perpendicular to each other and usually a line north and east as shown in this graphic. Note that point L is referenced to point K by a departure or easting of 594.03 feet in a northing or latitude of 143.89 feet. Northings and eastings are also called delta Y and delta X respectively. Y and X are the standard designators for north and east distances in all trigonometric calculation procedures. After looking at the two preceding graphics we can see that a polar measurement by bearing and distance and a rectangular measurement delta Y and delta X are functions of each other. In fact it can be said that every line measured by bearing and distance also has a corresponding delta X delta Y and delta X which can also be called latitude and departure. As shown in this graphic the latitude is a distance measured in north-south direction while departures are measured in north-south direction. Latitudes are a positive value when measured to the north and negative when measured south. Likewise departures are positive when measured to the east and negative when measured to the west. Therefore when computing latitudes and departures remember to assign the proper algebraic sign to distinguish which quadrant they fall into. After looking at the last couple of graphics it is evident that this dimension and the latitude and departure dimension when superimposed formulate triangle. This is what calculation of latitudes and departures from a survey line is based on. The process is almost identical to right triangle formulas. Almost identical because instead of dealing with angle A as we did earlier in the video we are now going to use the bearing or azimuth angle instead. This is a computing process because the latitude and departure will be reversed if angle A is used. To calculate a latitude and departure of a survey course simply multiply the length of the course by the cosine of the bearing or azimuth to determine latitude. Multiply length by the sign of the bearing or azimuth to find the departure. This graphic shows the equation format for latitude and departure computations. When using bearings it is necessary to select and assign the proper algebraic sign to the latitudes and departures. When using azimuths the algebraic signs of the sine and cosine functions will produce the correct algebraic signs for the computed latitudes and departures. Let's try a determination of latitude and departure as shown in this sketch. First find sine and cosine of the bearing. Sine equals 0.849218 and cosine equals 0.52043 then multiply the length 549.81 feet by the functions. This equals a latitude of plus 290.32 feet and a departure of plus 466.91 feet. Because it is a northeast bearing latitude and departure will have positive signs. Let's try the one shown in this sketch. The coarse bearing is south 46 degrees 11 minutes 57 seconds rest and distance is 576.33 feet. What is the latitude and departure? Stop the tape, calculate the values and then get back to me. Hello again. If you did the same work I did you would have gotten a latitude of negative 398.91 feet and a departure of negative 451.97 feet. Note that both values are negative since the course lies in a southwestern direction. A Clair's Traverse survey begins and ends at the same point forming a geometrical polygon. Therefore when algebraically summarized latitudes and departures should cancel each other out both sounds algebraically totaling to zero for the Traverse. This graphic shows a summary table for the Traverse. Note the algebraic signs and the totals would indicate misclosure. As I mentioned earlier the totals should always be zero but really are due to the random error always present in any and all survey measurements. Remember this graphic can compare it to the next one which is a sketch of a Traverse with the latitudes and departures shown plotted around the chorus. On this sketch notice the Traverse forming a Clair's five sided polygon that should but really will ever close perfectly. Note how the latitudes and departures shown around the Traverse appear to form a rectangle. After they were algebraically summarized as shown in the last graphic the error of closure is defined. The process of Traverse closure and adjustment and also the computation of area by double meridian distance method will be fully addressed later in another unit of this series of videos. When the latitude and departure between two unconnected survey points are known they can be used to determine the unknown bearing in distance between the points. This operation is called the inverse process and is based on the formulas shown in the next graphic. Note that in the inverse process latitude and departure will be called delta y for latitude and delta x for departure. The equations shown are used to determine the bearing in distance of the missing chorus. First the bearing AB. Tangent bearing or azimuth AB equals delta x AB divided by delta y AB. The distance AB is determined by the second equation. Distance AB equals delta x AB divided by sign of bearing AB. The results of the computation can be checked by application of the Pythagorean theorem. Distance AB equals the square root of delta delta y AB squared plus delta x AB squared. Before leaving latitudes and departures or delta y and delta x as I like to call them, let's do an inverse problem. From the data shown on this graphic we want to solve for the missing information for line R to S. To calculate the azimuth in line, length of line RS delta y equals minus 1436.11 feet and delta x equals 1421.18 feet. Therefore 1421.18 feet divided by negative 1436.11 feet equals arc tangent 0.989 604 equals 134 degrees 42 minutes and 2 seconds azimuth. 101,421.18 feet divided by the sign of the bearing sign is 0.703402 equals 2,020.44 feet. Check is the square root of negative 1436.11 squared plus 1421.18 feet squared equals 2,020.44 feet. Problem solved and proven. Let's move on to converting delta y's and delta x's into plane coordinates. Rectangular coordinates locate the positions of survey stations onto a perpendicular reference system having a single point of origin. This type of system is based on a pair of axes at right angles to each other. On this sketch the line labeled x-axis runs in an east-west direction. Distances measured in the easterly direction will have positive values. The y-axis runs north-south and distances measured from south to north will have a plus value. Notice that the point of origin has a coordinate value of 00. The point of origin of all systems should be located far enough to the west and south and assign this value. This assures that all points in the system will be located in the northeast quadrant of the system and have positive coordinate values. Coordinate systems may be a simple plane system such as we will be discussing for the next few minutes or can be a very complex statewide system such as the California coordinate system of 1983 that will be discussed in the later video unit. This graphic is included to illustrate how coordinate values for individual survey points are located. Let's take a look at point Q. Notice that it is located 75 feet north in 55 foot east of the origin. Therefore, its coordinate value or address will be y equals 75 feet and x equals 55 feet. Moving on up to the next point R, we see that its delta y is 173.04 feet and delta x is 154.56 feet. Its coordinates will be y equals 173.04 feet and x equals 154.56 feet. Now on the move northeast to lead to point S we find we have traveled a delta y of 83.08 feet and a delta x of 135.11 feet from point R. Adding these deltas to the coordinates for point R, we determine that point S will have coordinates of y equals 256.12 feet and x equals 289.67 feet. While we are still looking at this sketch, also notice the relationship of all points on the system to the origin and the relationship of each point on the system to any other point. For instance, a quick bit of simple math tells us that R is located 98.04 feet north in 99.56 feet east of Q by rectangular location. Some more quick math also tells us that R is also located north 45 degrees 26 minutes 26 seconds east 139.80 feet from Q by polar location. Let's leave this graphic and discuss this quick math in a little more detail in a bit. Coordinates for survey points are computed by algebraically adding the delta y and delta x of each succeeding course to the coordinate value of the starting point or the point at the beginning of the survey. This graphic illustrates the equations used to calculate coordinates of survey points starting from a coordinated survey station. Assume a coordinate is given for station R and given delta y and delta x you're asked to determine coordinates for station s. The equations for computing coordinates for station s are y of s equals y of r plus delta y rs x of s equals x of r plus delta x of rs. This next graphic presents an application of equations we have just looked at. A short while back in the presentation I showed you a graphic of latitudes in the departure tabulation that did not close very well. While behind the scenes I adjusted the misclosure out of the figure by a method you will share in one of the later videos of the series. In the graphic now on screen notice the delta y AB 270.17 has been added to the starting y coordinate for station A, 6000 feet. Making the y coordinate for station B 6,270.17 feet. The x coordinate for station B is found by adding delta x AB 64.96 feet to the starting x coordinate for station A. The x coordinate for B 64.96 feet. The next step is to determine coordinates of point C by adding delta y BC negative 68.97 feet to the y coordinate for station B 6,270.17 feet. The y coordinate for station C is found to be 6,201.20 feet. The x coordinate for station C is found by adding delta x BC 197.97 feet to the x coordinate for station B. The x coordinate for C is calculated at 2,362.93 feet. Repeating this operation the remaining way around the traverse we close back into A the starting point with a perfect closure. The equations just discussed though appearing in a slightly different form are identical with those used to determine polar dimensions that we discussed in detail in the latitudes and departure section of this video. Other than finding delta y and delta x by subtracting the respective coordinates the math is the same. Therefore I will not present any practice problems for this operation. Rest assured though that this operation will be covered in the workbook problems. Survey surveys are placed on or referenced to either a state plane coordinate system or a local arbitrary system. Locating survey points on a coordinate system have these distinct advantages. Location of all points are uniquely defined both on the ground and to each other. The survey can easily be plotted and mapped. Lost or destroyed points can be easily replaced. Data storage and retrieval is much more efficient and survey computations are greatly simplified. In this section I will extend the use of coordinates to find intersection points of two straight lines intersection points of a straight line in an arc and intersection points of two arcs. These type of situations are solved through the use of coordinate geometry which is rooted in analytic geometry. This section of the math video describes some of the basics of coordinate geometry along with some calculation examples. I do not intend to fairly explain coordinate geometry. I only wish to familiarize the viewer with the operations and software that is now being used to solve problems by coordinate geometry. The coordinate geometry equations for solution of missing elements are a little too complex to include on this video. However, that they are sufficiently addressed in the workbook. The first in a series of intersection problems involves finding the location where lines AC and BD intersect as shown in the upcoming graphics. Ideally, the answers we are looking for are the coordinate value of station E the distance of line AE and the distance of line BE. The given values that we will work with as shown in the graphic are coordinates of point A coordinates of point B the bearing of line AE and the bearing of line BE. The first step to take in the solution is to examine the given data to see which case the solution will fit into. Although we have already identified this as a bearing-bearing case let's look at the givens. We have a bearing for both lines A to E and B to E and can inverse between A and B to find either a base for an oblique triangle solution or enter the coordinates into a computer program to the bearing-bearing solution. In a typical programed computer solution coordinate values for A and B are entered along with the bearing of lines A to E and B to E. The computer will then process data using the coordinate geometry equations and give the coordinate value of point E along with the distance of lines AE and BE. To use the oblique triangle solution after inversing it is then necessary to determine all angles in the triangle from the given bearings in the one computed. Use the distance computed from the inverse as the base for an angle-angle side solution. The answers as shown on this graphic are the coordinate value of station E is Y 752.35 X 1240.83 feet. The distance of line AE is 415.51 feet. The distance of line BE is 477.76 feet. The next in the intersection of straight line series is finding the location where lines K to M and L to N intersect as shown on this next graphic. Again the answers we are looking for include the coordinate value of station O, the bearing of line AE and the distance of line BE. The given values we will work with as shown on the graphic are coordinates of point N, coordinates of point K, distance of line K to O, bearing of line N to O. Let's examine the given data to see what case the solution will fit into. From looking at the given values we have a bearing for line N to O and a distance for K to O. And it can inverse between N and K to find either a base for an oblique triangle solution or enter the coordinates O into a computer program for a bearing distance solution. In a typical program computer solution coordinate values for N and K are entered along with the bearing of lines N, O and the distance of line K, O. The computer will then process the data using the coordinate geometry equations. The coordinate value of point O and the distance for line N, O along with the bearing of line K, O will be displayed or the proper storage registers. This time in using the oblique triangle for a solution after inversing it is a little more involved as it is necessary to first determine the interior angle at O. This however will complete all angles in the triangle leading to an angle-angle-side solution. The answers for the bearing distance problem are as shown on the screen. The coordinate value of station O is Y 609.87 feet, X 695.75 feet, the distance of line N, O is 496.23 feet, the bearing of line K, O is north 25 degrees, 36 minutes, 44 seconds east. The last of the intersection of straight line series is to determine where the lines T, V and W, V intersect as shown in this next graphic. This time the answers being sought are the coordinate value of X, the bearing of line X to V, and the bearing of line X to U. The given values that we will work with this time are shown on the graphic as coordinates of point U, coordinates of point V, distance of line V to X, distance of line U to X. Let's look at what is given to decide which case this solution will fit into. From looking at the given values, we find that we have distances for both line V to X and line U to X. We can then inverse between V and U to find either a base for an oblique triangle solution or else enter the coordinates of V and U into a computer to solve this problem. For the computer solution, coordinate values for V and U are entered along with the distances of lines V to X and U to X. The computer will then process the data using the coordinate geometry equation for distance-distance solution, displaying the coordinate value of point X and the bearings for lines V to X and U to X. The answer to this problem will be based on the case for three given sides and can be solved with a variation of the distance-distance solution. The answers for the distance-distance problem are shown on this screen. The coordinate value of station X is Y 1652.57 and X 1733.15. The bearing of line U to X is North 7903.30 East. The three preceding intersection solutions have addressed all the cases possible with straight-line solutions. I must caution you that in the distance-distance case there are two solutions to the problem that are equally correct. Be sure that you always draw a sketch to guide you in the choice of the solution to fit your needs. Two other intersection situations, intersection of a straight line and arc and intersection of two arcs have not been addressed in this discussion. Solutions for these two cases are based on the same geometry and selection of formulas used for straight-line intersection problems. The viewer needs only to draw a sketch of the solution should become quite clear. The discussion of intersection problems has brought us to the end of this video presentation. Thank you for your attention throughout this long session. I sincerely hope that I was able to make basic surveying math a little less formidable and a little more friendlier. If nothing else, I offer you this advice. Always, always, always draw a sketch. Recalculations as simple as possible and be neat and meticulous in your notes. Good luck with your surveying career in general and with the land surveyors examinations processes in particular.