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The angle of elevation of the top of a tower as observed from a point in a horizontal plane through the foot of the tower is $32 ° .$ When the observer moves towards the tower at a distance of100 $m$ , he finds the angle of elevation of the top to be $63 °$ , then find the distance between the tower and the first position.

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146.7 m

145.7 m

143.7 m

149.7 m

answer is A.

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Detailed Solution

Given,
Angle of elevation of top of a tower is

32 ° .

Consider the figure,
Question Image

Let ‘h’ be the height of tower,

tan 63 ° = height of the tower base of the tower ⇒ tan 63 ° = h x ⇒ h = x tan 63 °

Similarly,

tan 32 ° = height of the tower distance of observer from tower ⇒ tan 32 ° = h x + 100 ⇒ h = (x + 100) tan 32 °

Comparing the heights,

x tan 63 ° = (x + 100) tan 32 ° ⇒ 19626 x = 0.6248 x + 62.48 ⇒ 13378 x = 62.48 ⇒ x = 46.7

The distance between the tower and first position,

d = x + 100 ⇒ d = 46.7 + 100 ⇒ d = 146.7 m

Therefore, the distance between the tower and first position is 146.7 m.
Hence, option 1 is correct.

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