The angle of elevation of the top pf a tower from two points of distances $a$ and $b$ metres from the base and in the same straight line with it are $α$ and $β$ respectively. Prove that the height of the tower is $h = \frac{(a - b) tan α tan β}{tan β - tan α}$

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Solution

R.E.F image
a = x + d
a = b + d
(d = a - b)
$t a n β = \frac{h}{x}$ $t a n α = \frac{h}{d + x}$
$∴ x = \frac{h}{t a n β}$
$∴ t a n α = \frac{h}{d + \frac{h}{t a n β}}$
$∴ d t a n α + h . \frac{t a n α}{t a n β} = h$
$∴ h = \frac{d t a n α}{1 - \frac{t a n α}{t a n β}} = \frac{d t a n . t a n β}{t a n β - t a n α}$
$= \frac{(a - b) t a n α . t a n β}{t a n β - t a n α}$

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