Question

help me out with sum plzzz it's urgent
will mark as brainliest ​

Answer 1

Let AB be the Tower of height 50m. Let CD be the pole of height X m. And Y be the distance between the tower and the pole.

In Triangle AEC

Tan 30°=AE/EC

1/√3 =50°-X/Y

Y =√3(50-X) ....(1)

In Triangle ABD

Tan 45°=AB/BD

1=50/Y

Y=50

√3(50-X)=50 .... {FROM (1)} 1.73(50-X)=50

86.5-1.73X=50

86.5-50=1.73X

36.5/1.73=X

X=21 (approx)

Hence height of pole=21m (approx)