From a 60 m high tower,the angles of depress | vedclass

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(D) Let $H = 60 \ m$ be the height of the tower and $h$ be the height of the house. Let $d$ be the distance between the tower and the house.From the g

Vedclass · Accepted Answer

$\cos \alpha \sin \beta$

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(D) Let $H = 60 \ m$ be the height of the tower and $h$ be the height of the house. Let $d$ be the distance between the tower and the house.From the geometry of the problem:$	an \alpha = \frac{H - h}{d} \Rightarrow d = \frac{60 - h}{	an \alpha}$$	an \beta = \frac{H}{d} \Rightarrow d = \frac{60}{	an \beta}$Equating the two expressions for $d$:$\frac{60 - h}{	an \alpha} = \frac{60}{	an \beta}$$(60 - h) 	an \beta = 60 	an \alpha$$60 	an \beta - h 	an \beta = 60 	an \alpha$$h 	an \beta = 60(	an \beta - 	an \alpha)$$h = 60 \left( \frac{\sin \beta}{\cos \beta} - \frac{\sin \alpha}{\cos \alpha} ight) \cdot \frac{\cos \beta}{\sin \beta}$$h = 60 \left( \frac{\sin \beta \cos \alpha - \cos \beta \sin \alpha}{\cos \beta \cos \alpha} ight) \cdot \frac{\cos \beta}{\sin \beta}$$h = \frac{60 \sin(\beta - \alpha)}{\cos \alpha \sin \beta}$Comparing this with the given expression $\frac{60 \sin(\beta - \alpha)}{x}$,we get $x = \cos \alpha \sin \beta$.

From a $60 \ m$ high tower,the angles of depression of the top and bottom of a house are $\alpha$ and $\beta$ respectively. If the height of the house is $\frac{60 \sin(\beta - \alpha)}{x}$,then $x =$

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