Question

the angle of elevation of the top of a tower from a point on the ground is 45° on walking 30 M towards the tower the angle of elevation becomes 60 degree find the height of the tower and the original distance from the foot of the tower​

Answer 1

here, here, ur ans

I hope it's correct

Answer 2

The height of the tower and the original distance from the foot of the tower both are $\frac{30\sqrt{3}}{\sqrt{3} - 1}$ meters.

Step-by-step explanation:

See the attached diagram.

Let the tower be AB with A as the foot. Initially, the angle of elevation of B from point C is 45°.

So, from Δ CAB, $\tan 45 ^{\circ} = \frac{AB}{CA} = \frac{H}{CA}$

⇒ $CA = \frac{H}{\tan 45^{\circ}} = H$ {Since tan 45° = 1}

Now, moving from C to D by 30 m, the angle of elevation becomes 60°.

So, from Δ DAB,

$\tan 60^{\circ} = \frac{H}{DA}$

⇒ $DA = \frac{H}{\tan 60^{\circ}} = \frac{H}{\sqrt{3}}$

Therefore, CA - DA = 30 m

⇒ $H - \frac{H}{\sqrt{3}} = 30$

⇒ $H = \frac{30\sqrt{3}}{\sqrt{3} - 1}$ meters.

And the original distance from the foot of the tower is $\frac{30\sqrt{3}}{\sqrt{3} - 1}$ meters.

(Answer)