Solution 1

In $\triangle A B C$ shown below, $\angle B A C=180^{\circ}-\angle A B C-\angle A C B=180^{\circ}-60^{\circ}-90^{\circ}=30^{\circ}$.

Since $\angle A D C$ is a straight angle, $\angle A D E=180^{\circ}-\angle C D E=180^{\circ}-48^{\circ}=132^{\circ}$.

In $\triangle A E D, \angle A E D=180^{\circ}-\angle A D E-\angle E A D=180^{\circ}-132^{\circ}-30^{\circ}=18^{\circ}$.

Since $\angle A E B$ is a straight angle, $\angle D E B=180^{\circ}-\angle A E D=180^{\circ}-18^{\circ}=162^{\circ}$.

Thus, the value of $x$ is 162 .

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-266.jpg?height=407&width=377&top_left_y=645&top_left_x=950)

## Solution 2

The sum of the interior angles of a quadrilateral is $360^{\circ}$.

In quadrilateral $B C D E$ shown above,

$$
\angle D E B=360^{\circ}-\angle E D C-\angle D C B-\angle C B E=360^{\circ}-48^{\circ}-90^{\circ}-60^{\circ}=162^{\circ}
$$

Thus, the value of $x$ is 162 .

ANSWER: (E)
In the diagram, the value of $x$ is
(A) 72
(B) 158
(C) 108
(D) 138
(E) 162

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-055.jpg?height=331&width=303&top_left_y=2201&top_left_x=1258)

Part B: Each correct answer is worth 6.