Question

The length of the shadow of a tower standing on level ground is found to be 2x metres longer when the sun's elevation is 30°than when it was 45°. The height of the tower in metres is

Options

• \[\left( \sqrt{3} + 1 \right) x\]

• \[\left( \sqrt{3} - 1 \right) x\]

• \[2\sqrt{3}x\]

• \[3\sqrt{2}x\]

Answer 1

Let h be the height of tower AB

Given that: angle of elevation of sun are`∠D=30°` and.`∠C=45°`

Then Distance`CD=2x`  and we assume `BC=x`Here, we have to find the height of tower.

So we use trigonometric ratios.

In a triangle,`ABC`

`⇒ tan C=(AB)/(BC)` 

`⇒ tan 45°=(AB)/(BC)` 

`⇒ 1=h/x`

`⇒ x=h` 

Again in a triangle ABD,

`⇒ tan D= (AB)/(BC+CD)` 

`⇒ tan 30°=h/( x+2x)`

`⇒1/sqrt3=h/(h+2x)`                  `[x=h]` 

`⇒ sqrt3h=h+2x` 

`h(sqrt3-1)=2x`

`⇒ h=2x/(sqrt3-1)`

⇒` h=(2x)/(sqrt3-1)xx(sqrt3+1)/(sqrt3+1)` 

`⇒ h=x(sqrt3+1)`