If the angles of elevation of a tower from two points distant a and b (a>b) from its foot and in the same straight line from it are 30° and 60°, then the height of the tower is

#### Options

\[\sqrt{a + b}\]

\[\sqrt{ab}\]

\[\sqrt{a - b}\]

\[\sqrt{\frac{a}{b}}\]

#### Solution

Let h be the height of tower AB

Given that: angle of elevation are`∠C=60°` and`∠D=60°` .`∠D=30°`

Distance `BC=b` and`BD=a`

Here, we have to find the height of tower.

So we use trigonometric ratios.

In a triangle `ABC`,

`⇒ tan C=( AB)/(BC)`

`⇒ tan 60°=(AB)/(BC)`

`⇒ tan 60°= h/b`

Again in a triangle *ABD*,

`⇒ tan D=(AB)/(BD)`

`⇒ tan 30°=h/a`

Again in a triangle *ABD*,

`⇒ tan D=(AB)/(BD)`

`⇒ tan 30°=h/a`

`⇒ tan (90°-60°)=h/a`

`⇒ cot 60°=h/a`

`⇒ 1/tan 60°=h/a`

`⇒ b/h=h/a` ` "put tan "60°=h/b`

`⇒ h^2=ab`

`⇒h=ab`

`h=sqrt(ab)`