From the bottom of a pole of height h, the an | vedclass

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(B) Let $H$ be the height of the tower and $d$ be the distance between the pole and the tower.From the right-angled triangle formed by the pole's bott

Vedclass · Accepted Answer

$\frac{h\cot(\alpha - \beta)}{\cot(\alpha - \beta) - \cot\alpha}$

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(B) Let $H$ be the height of the tower and $d$ be the distance between the pole and the tower.From the right-angled triangle formed by the pole's bottom and the tower's top,we have $	an\alpha = \frac{H}{d}$,which implies $d = H\cot\alpha$.From the right-angled triangle formed by the top of the pole and the top of the tower,we have $	an(\alpha - \beta) = \frac{H - h}{d}$,which implies $d = (H - h)\cot(\alpha - \beta)$.Equating the two expressions for $d$:$H\cot\alpha = (H - h)\cot(\alpha - \beta)$$H\cot\alpha = H\cot(\alpha - \beta) - h\cot(\alpha - \beta)$$h\cot(\alpha - \beta) = H\cot(\alpha - \beta) - H\cot\alpha$$h\cot(\alpha - \beta) = H(\cot(\alpha - \beta) - \cot\alpha)$$H = \frac{h\cot(\alpha - \beta)}{\cot(\alpha - \beta) - \cot\alpha}$

From the bottom of a pole of height $h,$ the angle of elevation of the top of a tower is $\alpha$ and the pole subtends an angle $\beta$ at the top of the tower. The height of the tower is

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