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Question
A tower stands vertically on the ground. From a point on the ground 20 m away from the foot of the tower, the angle of elevation of the top of the tower is 60°. What is the height of the tower?
Solution
Let AB be the tower of height, hm and C be the point on the ground, makes an angle of elevation 60° with the top of tower AB.
In a triangle ABC, given that BC = 20 m and ∠C = 60°
Now we have to find the height of tower AB, so we use trigonometrical ratios.
In the triangle ABC
`=> tan C = "Opposite side (AB)"/("Adjacent side (BC)")`
i.e `tan 60^@ = (AB)/20`
`=> AB = 20sqrt3`
`= 20 sqrt3`
∴ Height of tower H = `20sqrt3m`
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