The angle of elevation of the top of a T.V. t | vedclass

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(B) Let $OP = h$ be the height of the tower. Let $O$ be the foot of the tower.In $	riangle OAP$,$\angle OAP = \alpha$. So,$OA = h \cot \alpha$.In $

Vedclass · Accepted Answer

$a \sin 2\alpha$

Vedclass · Answer

(B) Let $OP = h$ be the height of the tower. Let $O$ be the foot of the tower.In $	riangle OAP$,$\angle OAP = \alpha$. So,$OA = h \cot \alpha$.In $	riangle OBP$,$\angle OBP = 2\alpha$. So,$OB = h \cot 2\alpha$.Given $AB = a$,we have $OA - OB = a$.Therefore,$h(\cot \alpha - \cot 2\alpha) = a$.$h \left( \frac{\cos \alpha}{\sin \alpha} - \frac{\cos 2\alpha}{\sin 2\alpha} ight) = a$$h \left( \frac{\sin 2\alpha \cos \alpha - \cos 2\alpha \sin \alpha}{\sin \alpha \sin 2\alpha} ight) = a$$h \left( \frac{\sin(2\alpha - \alpha)}{\sin \alpha \sin 2\alpha} ight) = a$$h \left( \frac{\sin \alpha}{\sin \alpha \sin 2\alpha} ight) = a$$h \left( \frac{1}{\sin 2\alpha} ight) = a$$h = a \sin 2\alpha$.

The angle of elevation of the top of a $T.V.$ tower from three points $A, B, C$ in a straight line through the foot of the tower are $\alpha, 2\alpha, 3\alpha$ respectively. If $AB = a$,the height of the tower is

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