In an art gallery, a $2 \mathrm{~m}$ high painting, $B T$, is mounted on a wall with its bottom edge $1 \mathrm{~m}$ above the floor. A spotlight is mounted at $S, 3 \mathrm{~m}$ out from the wall and $4 \mathrm{~m}$ above the floor. The size of $\angle T S B$ is closest to
(A) $27^{\circ}$
(B) $63^{\circ}$
(C) $34^{\circ}$
(D) $45^{\circ}$
(E) $18^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-124.jpg?height=257&width=268&top_left_y=1709&top_left_x=1362)
Let $X$ be the point on $S F$ so that $B X$ is perpendicular to $S F$.

Then $B X=3, X F=1$ and $X S=3$.

Therefore, $\triangle B X S$ is an isosceles right-angled triangle, and so $\angle B S X=45^{\circ}$.

Let $Y$ be the point on $S F$ so that $T Y$ is perpendicular to SF.

Then $T Y=3$ and $S Y=1$, and $\triangle S T Y$ is right-angled.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-242.jpg?height=349&width=284&top_left_y=1907&top_left_x=1365)

Therefore, $\tan (\angle T S Y)=\frac{3}{1}=3$ and so $\angle T S Y \approx 71.6^{\circ}$.

Thus, $\angle T S B=\angle T S Y-\angle B S F \approx 71.6^{\circ}-45^{\circ}=26.6^{\circ}$, which is closest to $27^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-243.jpg?height=349&width=271&top_left_y=243&top_left_x=1363)

ANSWER: (A)