Question

Bridge across a valley is h metres long.
There is a temple in the valley directly below the bridge.
The angle of depression of the top of the temple from the two end of the bridge have measure
$\alpha$
and
$\beta$
prove that the height of the bridge above the top of the temple is
$h( \tan \alpha \times \tan \beta ) \div \\ (\tan \alpha + \tan \beta )$
​

Answer 1

height of the bridge above the top of the temple  =  h (TanαTanβ) / (Tanα + Tanβ)

Step-by-step explanation:

Let say Temple as at distance of a from the end of bridge having angle of depression = α    &

at distance b  from the end of bridge having angle of depression = β

Let say height of the bridge above the top of the temple is = x

The  Tanα = x /a

=> a = x/Tanα

        Tanβ = x/b

=> b = x/Tanβ

h = a + b

=> h   = x/Tanα  + x/Tanβ

=> h = x (Tanβ + Tanα) / TanαTanβ

=> x = h (TanαTanβ) / (Tanβ + Tanα)

=> x =  h (TanαTanβ) / (Tanα + Tanβ)

Proved

QED

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