A tower of height b subtends an angle alpha a | vedclass

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(B) Let the tower be $AB$ with height $b$ and the pole be $BP$ with height $p$. The point $O$ is at distance $a$ from $A$. Given $\angle AOB = \alpha$

Vedclass · Accepted Answer

$b \left( \frac{a^2 + b^2}{a^2 - b^2} \right)$

Vedclass · Answer

(B) Let the tower be $AB$ with height $b$ and the pole be $BP$ with height $p$. The point $O$ is at distance $a$ from $A$. Given $\angle AOB = \alpha$ and $\angle POB = \alpha$. In $	riangle OAB$,$	an \alpha = \frac{AB}{OA} = \frac{b}{a}$. In $	riangle OAP$,$\angle POA = 2\alpha$. So,$	an 2\alpha = \frac{AP}{OA} = \frac{p + b}{a}$. Using the formula $	an 2\alpha = \frac{2 	an \alpha}{1 - 	an^2 \alpha}$,we get: $\frac{2(b/a)}{1 - (b/a)^2} = \frac{p + b}{a}$ $\frac{2b/a}{(a^2 - b^2)/a^2} = \frac{p + b}{a}$ $\frac{2ba}{a^2 - b^2} = \frac{p + b}{a}$ $p + b = \frac{2ba^2}{a^2 - b^2}$ $p = \frac{2ba^2}{a^2 - b^2} - b$ $p = \frac{2ba^2 - b(a^2 - b^2)}{a^2 - b^2}$ $p = \frac{2ba^2 - ba^2 + b^3}{a^2 - b^2}$ $p = \frac{ba^2 + b^3}{a^2 - b^2} = \frac{b(a^2 + b^2)}{a^2 - b^2}$.

$A$ tower of height $b$ subtends an angle $\alpha$ at a point $O$ on the level of the foot of the tower and at a distance $a$ from the foot of the tower. If a pole of height $p$ mounted on the tower also subtends an equal angle $\alpha$ at $O$,the height of the pole $p$ is:

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