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I tried to solve this Trigonometry question, but I do not know how to solve. I read that the

circle has radius 1 and center at (0.0) as the unit circle is plotted in the coordinate

system. I should record the angles $v$ and $w$ in unit circle so that the following are met:

$v$ is an acute angle with $\sin(v) = 0.9$ and w is an obtuse angle with $\sin(w) = 0.9$

You can see the circle here

i don't know how to solve it, but is it something like : $(\sin(0.9) * 100) - \pi = 75.1910983$ ? please help me out, give me examples and information

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That the sine is positive means you can restrict to the first and second quadrants (why?); the angle on one quadrant is acute and the angle on the other quadrant is obtuse. Now, draw a picture and report back. –  Ｊ. Ｍ. Feb 5 '12 at 16:06
I don't get it, please give me example –  user1022734 Feb 5 '12 at 16:11

2 Answers 2

up vote 1 down vote accepted

I think you simply need to find the points on the unit circle with $y$-coordinate $.9$ (recall $\sin\theta$ is the $y$ coordinate of the appropriate point on the unit circle).

For the acute angle, $v$, with $\sin v=.9$, $v$ is in the first quadrant. The $y$ coordinate is $.9$, and using the Pythagorean Theorem, the $x$-coordinate is $x=\sqrt{1-.9^2}=\sqrt{19/100}={1\over10}\sqrt{19}$.

enter image description here

The obtuse angle is in the second quadrant with $y$-coordinate $.9$. You can solve for the $x$-coordinate as above.

I'm not sure that you need explicitly find the measure of the angle; but if so, take the inverse sin of the angles.

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Instead of solving for the $x$ coordinate, I think the OP is just supposed to draw the horizontal line $y=0.9$ and see where it intersects his (compass-drawn) circle. Finding the angle measure is almost certainly not part of the intended solution. –  Henning Makholm Feb 5 '12 at 16:28
Henning Makholm, det korrekt, men kan du hjælpe mig? sin-1(0,9) = 64.1580672368 º –  user1022734 Feb 5 '12 at 16:32
@user1022734 (who said "that correct, but can you help me?"): It is not clear to me what more help you need. As far as I can see, David's answer shows you in detail what you need to do. –  Henning Makholm Feb 5 '12 at 16:34
sin-1(0,9) = 64,16º sin(180º-V) = sin(V) = 115,84º , but i don't know how to draw it... –  user1022734 Feb 5 '12 at 16:37
@user1022734: Can you see the drawing in the answer? –  Henning Makholm Feb 5 '12 at 16:43

Essentially, you're looking for an angle $v$ that satisfies: $$ \sin(v) = 0.9 \;\;\; 0\leq v \leq \frac{\pi}{2}$$ (angles in rad) an angle $w$ that satisfies: $$ \sin(w) = 0.9 \;\;\; \frac{\pi}{2}\leq w\leq \pi$$ The value for $v$ follows easily from the first equation; just take the inverse sine of both sides. For the second equation, recall that $\sin(\pi - \theta) = \sin(\theta)$. Therefore, $w = \pi - v$

Note that this is a purely algebraic approach, and ignores the unit circle, which may well be the point of the exercise. I refer you to David's answer for a better explanation of the geometric approach.

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