The angle of elevation of the top of a TV tow | vedclass

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

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$a \sin 2 \alpha$

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(C) Let $OP = h$ be the height of the tower,where $O$ is the foot of the tower.In $	riangle OAP$,$\angle OAP = \alpha$. Thus,$OA = h \cot \alpha$.In $	riangle OBP$,$\angle OBP = 2 \alpha$. Thus,$OB = h \cot 2 \alpha$.Given $AB = a$,we have $OA - OB = a$.Therefore,$h \cot \alpha - h \cot 2 \alpha = a$.$h (\frac{\cos \alpha}{\sin \alpha} - \frac{\cos 2 \alpha}{\sin 2 \alpha}) = a$.$h (\frac{\sin 2 \alpha \cos \alpha - \cos 2 \alpha \sin \alpha}{\sin \alpha \sin 2 \alpha}) = a$.Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we get:$h (\frac{\sin(2 \alpha - \alpha)}{\sin \alpha \sin 2 \alpha}) = a$.$h (\frac{\sin \alpha}{\sin \alpha \sin 2 \alpha}) = a$.$h (\frac{1}{\sin 2 \alpha}) = a$.$h = a \sin 2 \alpha$.Thus,the height of the tower is $a \sin 2 \alpha$.

The angle of elevation of the top of a $TV$ 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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