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 have measures α and B. Prove that the height of the bridge above the top of the temple is (tanα.tanß)/tanα+tanß*m.

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,

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

i.e. $\tan \alpha=\frac{y}{x}$   ...............(1)

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

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

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

i.e. $x\tan \alpha=(h-x)\tan \beta$

i.e. $x\tan \alpha=h\tan \beta-x\tan \beta$

i.e. $x(\tan \alpha+\tan \beta)=h\tan \beta$

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

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

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

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

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

Hence proved.

Answer 2

"Answer: $y = (tan\alpha.tan\beta)/ tan\alpha +tan\beta \times m$

let y = vertical distance between temple and bridge

              H = length of the bridge

              X = horizontal distance between temple and one end of the bridge

From triangle OAB

Tanα = (y/x) --------------- (1)

From triangle OPQ

$Tan\beta = (y/(h-x))$ ------------(2)

Solving (1) and (2) we get

$tan\alpha/tan\beta = (y\times (h-x))/ (x\times y)$

$h Tan\beta - x Tan\beta = y Tan\alpha$

$x = (h Tan\beta - y\times tan\alpha)/tan\beta$

by substituting x value in equation (1) we get

$y = (tan\alpha.tan\beta)/ tan\alpha +tan\beta \times m$"