Let $t$ be TNYWR.

In the diagram, $A B C D$ is a trapezoid with $A B$ parallel to $D C$ and $B C=B D$. If $\angle D A B=x^{\circ}$, $\angle A D B=18^{\circ}$, and $\angle D B C=6 t^{\circ}$, what is the value of $x$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_24bfa07284b5df794287g-1.jpg?height=382&width=634&top_left_y=1408&top_left_x=735)

##
The triangle with an angle of $6 t^{\circ}$ is isosceles, so each of the other angles measures

$$
\frac{1}{2}\left(180^{\circ}-6 t^{\circ}\right)=90^{\circ}-3 t^{\circ}
$$

Since $A B$ is parallel to $D C$, we have that $\angle A B D=90^{\circ}-3 t^{\circ}$.

The angles in $\triangle A B D$ sum to $180^{\circ}$, so $x+18+(90-3 t)=180$ which can be solved for $x$ to get $x=180-90-18+3 t=72+3 t$.

Since $t=7$, we can substitute to get $x=72+3(7)=93$.