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Question
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Solution
Let AB be the tower and the angle of elevation from point C (on ground) is 30°.
In ΔABC,
`("AB")/("BC") = tan 30°`
`("AB")/(30) = 1/sqrt3`
AB = `(30)/sqrt3`
AB = `(30)/sqrt3 xxsqrt3/sqrt3`
AB = `10sqrt3` m
Therefore, the height of the tower is `10sqrt3` m.
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