A bridge across a valley is $h$ metres long. There is a temple in the valley directly below the bridge. The angles of depression of the temple from the two ends of the bridge have measures $α$ and $β .$ Prove that the height of the bridge above the top of the temple is $\frac{h (tan α . tan β)}{tan α + tan β} m .$

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Solution

We have

Let $A P = h$

As shown in the fig. let $D$ be the position of the temple.

According to the question,

$∠ P A D = α$

$∠ A P D = β$

Draw a line

$A B = P Q = y$

Let $B D = x$ then $D Q = h - x$

Then,

$I n Δ D A B$

$tan α = \frac{A B}{B D}$

$tan α = \frac{y}{x} . . . . . . . (1)$

$I n Δ D P Q$

$tan β = \frac{P Q}{D Q}$

$tan β = \frac{y}{h - x} . . . . . . . (2)$

Dividing equation $(2)$ by $(1)$ and we get,

$\frac{tan β}{tan α} = \frac{y}{h - x} \times \frac{x}{y}$

$\frac{tan β}{tan α} = \frac{x}{h - x}$

$h tan β - x t a n β = x t a n α$

$h tan β = x t a n α + x t a n β$

$x (tan α + tan β) = h tan β$

$x = \frac{h tan β}{(tan α + tan β)}$

Put the value of x in equation $(1)$ and we get,

$y = \frac{h tan α tan β}{(tan α + tan β)}$

Hence, the bridge above the top of temple is $\frac{h tan α tan β}{(tan α + tan β)}$ mtr.

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A bridge across a valley is h metres long. There is a temple in the valley directly below the bridge. The angles of depression of the temple from the two ends of the bridge have measures α and β. Prove that the height of the bridge above the top of the temple is h(tanα.tanβ)tanα+tanβ m.