Question

A TV tower stands vertically on a bank of a canal, with a height of10✓13 m. From a point on the
other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From
another point on the line joining this point to the foot of the tower, the angle of elevation of the top
of the tower is 30°. Find the distance between the
opposite bank of the canal and the point with 30°
angle of elevation.​

Answer 1

Answer:

let BC= y metre ,DB= x metre

hieght of tower =10√3m

Step-by-step explanation:

In ∆ABD,

AB/DB=tan30°

10√3/x=1/√3

x =10√3×√3

x=10×3,

x=30

now, In ∆ABC

AB/BC=tan60°

10√3/y=√3

10√3=y√3

10√3/√3=y

now we have to find DC

therefore, DB-CB =DC

30-10√3/√3=DC

DC=20m ans.