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Question
The angle of elevation of the top of a tower, from a point on the ground and at a distance of 160 m from its foot, is found to be 60°. Find the height of the tower.
Solution
Let AB be the tower and C is the point which is 160 m away from the foot of the tower, i.e CB = 160 m
Let height of the tower be h
Now in right ΔABC, we have
`tan theta = (AB)/(BC)`
`=> tan 60^circ = h/160`
∴ h = 160 × tan 60°
= `160 xx sqrt(3) m`
= 160 × (1.732) m
= 277.12 m
∴ Required height of the tower = 277.12 m
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