The angles of elevation of the top of a tower | vedclass

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(A) Let $AB$ be the tower of height $h$ units.Let the points $P$ and $Q$ be at distances $PB = a$ and $QB = b$ from the base $B$ of the tower.Given th

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$\sqrt{ab}$

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(A) Let $AB$ be the tower of height $h$ units.Let the points $P$ and $Q$ be at distances $PB = a$ and $QB = b$ from the base $B$ of the tower.Given that the angles of elevation are complementary,let $\angle AQB = 	heta$ and $\angle APB = 90^{\circ} - 	heta$.In $	riangle AQB$,$	an 	heta = \frac{AB}{QB} = \frac{h}{b}$.In $	riangle APB$,$	an(90^{\circ} - 	heta) = \frac{AB}{PB} = \frac{h}{a}$.Since $	an(90^{\circ} - 	heta) = \cot 	heta$,we have $\cot 	heta = \frac{h}{a}$.Multiplying the two equations: $	an 	heta \cdot \cot 	heta = \left(\frac{h}{b}ight) \cdot \left(\frac{h}{a}ight)$.Since $	an 	heta \cdot \cot 	heta = 1$,we get $1 = \frac{h^2}{ab}$.Therefore,$h^2 = ab$,which implies $h = \sqrt{ab}$.

The angles of elevation of the top of a tower from the points $P$ and $Q,$ at distances of $a$ and $b$ respectively from the base of the tower and in the same straight line with it,are complementary. The height of the tower is

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