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The Angles of Elevation of the Top of a Rock from the Top and Foot of a 100 M High Tower Are Respectively 30° and 45°. Find the Height of the Rock. - Mathematics

The angles of elevation of the top of a rock from the top and foot of a 100 m high tower are respectively 30° and 45°. Find the height of the rock.

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Solution

Let AB be the height of Rock which is Hm. and makes an angle of elevations 45° and 30° respectively from the bottom and top of the tower whose height is 100 m.

Let AE = hm, BC = x m and CD = 100.

∠ACB = 45°, ∠ADE = 30°

We have to find the height of the rock

We have the corresponding figure as

So we use trigonometric ratios.

In ΔABC

`tan 45^@ = (AB)/(BC)`

`=> 1- (100 + h)/x`

`=> x = 100 + h`

Again in Δ ADE

`tan 30^@ = (AE)/(DE)`

`=> 1/sqrt3 = h/x`

`=> 100 + h = sqrt3h`

=> h = 136.65

H = 100 + 136.65

=> H = 236.65

Hence the height of rock is  236.65 m

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APPEARS IN

RD Sharma Class 10 Maths
Chapter 12 Trigonometry
Exercise 12.1 | Q 64 | Page 34
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