Problem Statement
In an art gallery, a 2 m high painting, $BT$, is mounted on a wall with its bottom edge 1 m above the floor. A spotlight is mounted at $S$, 3 m out from the wall and 4 m above the floor. The size of $\angle TSB$ is closest to
Problem Link
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Solution
$$\text{Connect B to SF at point A} \implies AB = 3, \angle BAS = 90^{\circ}$$ $$\therefore BS^2 = AB^2 + AS^2 \implies BS = 3\sqrt{2}$$ $$\text{Since } \angle TBA = 90^{\circ} \text{ and } \angle SBA = 45^{\circ} \implies \angle TBS = 45^{\circ}$$ $$\text{Using cosine law: } ST^2 = 2^2 + (3\sqrt{2})^2 - 2(2)(3\sqrt{2})(sin(45^{\circ}))$$ $$\therefore ST = 10$$ $$\text{Using sine law: } \dfrac{sin(\angle TSB)}{TB} = \dfrac{sin(\angle TBS)}{TS}$$ $$\dfrac{sin(\angle TSB)}{2} = \dfrac{sin(45^{\circ})}{\sqrt{10}}$$ $$sin(\angle TSB) = \dfrac{1}{\sqrt{5}}$$ $$\angle TSB \approx 27^{\circ}$$ 
