Question

The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point, 40

m vertically above the initial point, the angle of elevation of the top of the tower is 45°. Find the height of the

tower. [Use 3 1.732 = ]​

Answer 1

Answer:

solution

 ΔYRQ, we have

tan45o=YRQR

1=YRx

YR=x or XP=x [because YR=XP] ---- (1)

Now In ΔXPQ we have

tan60o=PXPQ

3=xx+40 (from equation 1)

x(3−1)=40

x=3−140

x=1.73−140=54.79 m

So, height of the tower, PQ=x+40=54.79+40=94.79 m.

Distance PX=54.79 m.