By opposite angles, $\angle B A C=t^{\circ}$. This means $\angle B C A=180^{\circ}-2 t^{\circ}-t^{\circ}=180^{\circ}-3 t^{\circ}$. Again by opposite angles, $\angle D C E=\angle B C A$, so

$$
\begin{aligned}
x^{\circ} & =180^{\circ}-\angle C D E-\angle D C E \\
& =180^{\circ}-90^{\circ}-\left(180^{\circ}-3 t^{\circ}\right) \\
& =3 t^{\circ}-90^{\circ} .
\end{aligned}
$$

The answer to part (b) is 40 , so $t=40$ and $x=3 t-90=3 \times 40-90=30$.
Let $t$ be TNYWR.

Four line segments intersect in points $A, B, C, D$, and $E$, as shown. The measure of $\angle C E D$ is $x^{\circ}$. What is the value of $x$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_9c5b2a5bb2c0ebd3406fg-4.jpg?height=439&width=829&top_left_y=1447&top_left_x=651)

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