Question

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 top of the temple from the two
ends of the bridge area and B. Prove that the height of the bridge above the top
hſtan a tan B)
of the temple is
tan a +tan p
6 metres.♡
tonk full of water is emntied hy a pipe at the rate of 3
litres​

Answer 1

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.