A rectangle is drawn inside $\triangle A B C$, as shown. If $\angle B W Z=22^{\circ}$ and $\angle C X Y=65^{\circ}$, then the size of $\angle B A C$ is
(A) $87^{\circ}$
(B) $90^{\circ}$
(C) $93^{\circ}$
(D) $104^{\circ}$
(E) $82^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-104.jpg?height=385&width=583&top_left_y=1935&top_left_x=1118)
Solution 1

Since $W X Y Z$ is a rectangle, then $\angle Z W X=\angle W X Y=90^{\circ}$.

Therefore, $\angle A W X=180^{\circ}-\angle Z W X-\angle B W Z=180^{\circ}-90^{\circ}-22^{\circ}=68^{\circ}$.

Also, $\angle A X W=180^{\circ}-\angle W X Y-\angle C X Y=180^{\circ}-90^{\circ}-65^{\circ}=25^{\circ}$.

So, looking at $\triangle A W X, \angle B A C=\angle W A X=180^{\circ}-\angle A W X-\angle A X W=180^{\circ}-68^{\circ}-25^{\circ}=87^{\circ}$.

## Solution 2

Since $W X Y Z$ is a rectangle, then $\angle W Z Y=\angle X Y Z=90^{\circ}$, so $\triangle W Z B$ and $\triangle X Y C$ are both right-angled.

Thus, $\angle W B Z=90^{\circ}-\angle B W Z=90^{\circ}-22^{\circ}=68^{\circ}$ and $\angle X C Y=90^{\circ}-\angle C X Y=90^{\circ}-65^{\circ}=25^{\circ}$. Looking at $\triangle A B C, \angle B A C=180^{\circ}-\angle A B C-\angle A C B=180^{\circ}-68^{\circ}-25^{\circ}=87^{\circ}$.

## Solution 3

Draw a perpendicular from $A$ to $H$ on $B C$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-194.jpg?height=385&width=591&top_left_y=306&top_left_x=824)

Then $W Z, A H$ and $X Y$ are all parallel since each is perpendicular to $B C$, so $\angle B A H=\angle B W Z=22^{\circ}$ and $\angle C A H=\angle C X Y=\angle 65^{\circ}$.

Therefore, $\angle B A C=22^{\circ}+65^{\circ}=87^{\circ}$.

ANSWER: (A)