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Question

The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is 30° On advancing 150 m towards the foot of the tower, the angle of elevation becomes 60°. Show that the height of the tower is 129.9 metres.

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Solution

Let AB be the tower.
We have:
CD= 150 m, ∠ACB = 30o and ∠ADB = 60o
Let:
AB = h m and BD = x m

In the right ∆ABD, we have:
ABAD = tan 60o = 3

hx= 3
x = h3
Now, in the right ∆ACB, we have:
ABBC = tan 30o = 13

hx + 150 = 13
3h = x + 150

On putting x = h3 in the above equation, we get:
3h = h3 + 150
3h = h + 1503
2h = 1503
h =15032 = 753 = 75 × 1.732 = 129.9 m
Hence, the height of the tower is 129.9 m.

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