sir this is the important question from the book -ML aggarwal( avichal publication) class 10th , chapter20 , heights and distances
A vertical pole and a vertical tower are on the same level ground.
From the top of the pole the angle of elevation of the top of the tower is 60
and the angle of depression of the foot of the tower is 30.
Find the height of the tower if the height of the pole is 20 m.
question no 34 , heights and distances , ICSE board, ML Aggarwal
Consider TR as the tower
PL as the pole on the same level
Ground PL = 20 m
From the point P construct PQ parallel to LR
∠TPQ = 600 and ∠QPR = 300
Here
∠PRL = ∠QPR = 300 which are the alternate angles
Take LR = x and TR = h
TQ = TR – QR = (h – 20) m
In right triangle PRL
tan θ = PL/LR
Substituting the values
tan 300 = 20/x
So we get
1/√3 = 20/x
x = 20√3 m
In right triangle PQT
tan 600 = TQ/PQ
Substituting the values
√3 = (h – 20)/ x
√3 = (h – 20)/ 20√3
By cross multiplication
20√3 × √3 = h – 20
20 × 3 = h – 20
h = 60 + 20 = 80 m
Hence, the height of the tower is 80 m.