Question

From the top of a 120m high tower, a man observes two cars on the opposite sides of the tower and in straight line with the base of tower with angle of depression as 60° and 45° respectively. Find the distance b/w two cars.
Given that, √3 = 1.732

Answer 1

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Answer 2

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let AB represents the height of the tower
And two cars be at positions C and D

So CD represents the distance between the two cars

Given, height of the tower AB = 120 mts

Angle of cars in straight line with the base of tower with angle of depression = 60° and 45°

In $\triangle{ABC}$

Since $\angle{ACB}\: and\: \angle{XAC}$ are Alternate Angles, we have
$\angle{ACB} = \angle{XAC}=60^{\circ}$

In $\triangle{BAD}$

Since $\angle{BAD}\: and\: \angle{DAY}$ are Alternate Angles, we have
$\angle{BDA} = \angle{DAY}=45^{\circ}$

we have,
$tan = \frac{Opposite}{Adjacent}$

From $\triangle{ABC}$
$tan 60^{\circ} = \frac{AB}{CB}$

=> $\sqrt{3} = \frac{120}{CB}$

=> $CB = \frac{120}{\sqrt{3}}$

=> $CB = \frac{120}{\sqrt{3}}× \frac{\sqrt{3}}{\sqrt{3}}$

=> $CB = 40\sqrt{3}\: mts$

From $\triangle{BAD}$
$tan 45^{\circ} = \frac{AB}{BD}$

=> $1 = \frac{120}{BD}$

=> $BD = 120\: mts$

We have,
$CD = CB + BD$
=> $CD = 40\sqrt{3} + 120$

=> $CD = (40×1.732) + 120$

=> $CD = 69.28 + 120$

=> $CD = 189.28\: mts$

Therefore,
Distance between the two cars = 189.28 mts.