Question

from a point on the ground the angle of elevation of the top of a tower is observed to be 60degree . From a point 40m vertically above first point of observation the angle of elevation of the top of the tower is 30degree.Find the height of the tower and as horizontal distance from the point of observation

Answer 1

tan 30° = AB/BE.
1/√3 = AB/BE
BE=x√3

tan 60°= AC/CD
√3 = x + 40/CD
CD√3 = x +40
3x = x + 40
3x - x = 40
2x = 40
x = 20

So height of the tower is 60m and the horizontal distance is 20√3.




Hope it helps...

Answer 2

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.