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

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(C) Let $ED = h$ be the height of the tower. Points $A, B, C, D$ are collinear where $D$ is the foot of the tower.In $	riangle ADE$,$\angle EAD = \al

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$a \sin 2 \alpha$

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(C) Let $ED = h$ be the height of the tower. Points $A, B, C, D$ are collinear where $D$ is the foot of the tower.In $	riangle ADE$,$\angle EAD = \alpha$,$\angle EBD = 2\alpha$,$\angle ECD = 3\alpha$.In $	riangle ABE$,$\angle EAB = \alpha$ and $\angle EBA = 180^{\circ} - 2\alpha$.Thus,$\angle AEB = 180^{\circ} - (\alpha + 180^{\circ} - 2\alpha) = \alpha$.Since $\angle EAB = \angle AEB = \alpha$,$	riangle ABE$ is isosceles with $BE = AB = a$.In $	riangle BCE$,by the sine rule:$\frac{BE}{\sin 3\alpha} = \frac{CE}{\sin(180^{\circ} - 2\alpha)}$$\frac{a}{\sin 3\alpha} = \frac{CE}{\sin 2\alpha} \Rightarrow CE = \frac{a \sin 2\alpha}{\sin 3\alpha}$.In $	riangle CDE$,$\sin 3\alpha = \frac{h}{CE}$.$h = CE \sin 3\alpha = \left(\frac{a \sin 2\alpha}{\sin 3\alpha}ight) \sin 3\alpha = a \sin 2\alpha$.

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

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