The angle of elevation of the top of a tower as observed form a point in a horizontal plane through the foot of the tower is 32°. When the observer moves towards the tower a distance of 100 m, he finds the angle of elevation of the top to be 63°. Find the height of the tower and the distance of the first position from the tower. [Take tan 32° = 0.6248 and tan 63° = 1.9626]

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Solution

Let h be height of tower and the angle of elevation as observed from the foot of tower is 32° and observed move towards the tower with distance 100 m then angle of elevation becomes 63°.

Let BC = x and CD = 100

Now we have to find height of tower

So we use trigonometrical ratios.

In a triangle ABC,

Again in a triangle,

So distance of the first position from the tower is = 100 + 46.7 = 146.7 m

Hence the height of tower is m and the desires distance is m.

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