Question

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Answer 1

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Answer 2

Answer:

We are given that,

Length of the bridge = h meters

Angle of depression to the top of the temple from one side = α°

Angle of depression to the top of the temple from other side = β°

Since, 'Angle of depression and the corresponding angle of elevations have the same measure'

So, the angle of elevation made by the temple are α° and β°.

Let, the height of the bridge = y meters

Let, the horizontal distance to the temple from the base of the bridge = x meters, as shown in the figure.

Then, using trigonometric rule for the angles, we have,

\tan \alpha=\frac{Perpendicular}{Base}tanα=

Base

Perpendicular

i.e. \tan \alpha=\frac{y}{x}tanα=

x

y

...............(1)

and \tan \beta=\frac{y}{h-x}tanβ=

h−x

y

.................(2)

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

\frac{\tan \alpha}{\tan \beta}=\frac{h-x}{x}

tanβ

tanα

=

x

h−x

i.e. x\tan \alpha=(h-x)\tan \betaxtanα=(h−x)tanβ

i.e. x\tan \alpha=h\tan \beta-x\tan \betaxtanα=htanβ−xtanβ

i.e. x(\tan \alpha+\tan \beta)=h\tan \betax(tanα+tanβ)=htanβ

i.e. x=\frac{h\tan \beta}{\tan \alpha+\tan \beta}x=

tanα+tanβ

htanβ

Substitute the value of x in equation (1), we get,

\tan \alpha=\frac{y}{\frac{h\tan \beta}{\tan \alpha+\tan \beta}}tanα=

tanα+tanβ

htanβ

y

i.e. \tan \alpha=\frac{y\tan \alpha+\tan \beta}{h\tan \beta}tanα=

htanβ

ytanα+tanβ

i.e. y=\frac{h\tan \alpha\tan \beta}{\tan \alpha+\tan \beta}y=

tanα+tanβ

htanαtanβ

Hence proved.