Two parallel towers A and B of different heig | vedclass

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Question

(A) Let $d$ be the distance between the towers.From point $Q$ at height $10\,m$ on tower $A$,the angle of elevation to point $P$ at height $20\,m$ on

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) Let $d$ be the distance between the towers.From point $Q$ at height $10\,m$ on tower $A$,the angle of elevation to point $P$ at height $20\,m$ on tower $B$ is $	heta$.Thus,$	an 	heta = \frac{20 - 10}{d} = \frac{10}{d} \Rightarrow d = 10 \cot 	heta$.From point $P$ at height $20\,m$ on tower $B$,the angle of elevation to point $R$ at height $50\,m$ on tower $A$ is $2	heta$.Thus,$	an 2	heta = \frac{50 - 20}{d} = \frac{30}{d} \Rightarrow d = 30 \cot 2	heta$.Equating the two expressions for $d$:$10 \cot 	heta = 30 \cot 2	heta$$\cot 	heta = 3 \cot 2	heta$$\cot 	heta = 3 \left( \frac{\cot^2 	heta - 1}{2 \cot 	heta} ight)$$2 \cot^2 	heta = 3 \cot^2 	heta - 3$$\cot^2 	heta = 3 \Rightarrow \cot 	heta = \sqrt{3}$Therefore,$	heta = 30^{\circ}$.

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