Question

the angle of elevation of the top of a tower from a point on the ground is 60degrees. at a point 40m vertically above the first point of observation, the angle elevation is 30degrees.
find height of the tower and the distance from first point of observation to the tower
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Answer 1

Tan 30 = AB/BE

1/√3 = x/BE

BE = x√3m

Tan 60 = AC/CD

√3 = x + 40/CD

CD√3 = x + 40

Since BE is parallel to CD

x√3 x √3 = x + 40

3x = x + 40

3x - x = 40

2x = 40

x = 20

The height of the tower = x + 40

20 + 40 = 60m

The horizontal distance from the point of observation = x√3 = 20√3m

Hope it helps.