Question

A TV tower stands vertically on the side of a road. From a point on the other side directly opposite to the tower, the angle of elevation of the top of tower is 60°. From another point 10 m away from this 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 height of the tower and the width of the road.

Answer 1

Let the height of the tower = h mts say 

Width of the road be = x m. 

Distance between two points of observation = 10 cm. 

Angles of elevation from the two points = 60° and 30°. 

From the figure 

tan 60° = \(\frac{h}{x}\)

 √3 = \(\frac{h}{x}\)

⇒ h = √3x …….(1)

Also tan 30° = \(\frac{h}{x+10}\)

⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{x+10}\)

⇒ h = \(\frac{10+x}{\sqrt{3}}\) ………(2) 

From equations (1) and (2) h 

h = √3x = \(\frac{10+x}{\sqrt{3}}\)

∴ √3x = \(\frac{10+x}{\sqrt{3}}\)

√3 × √3x = 10 + x 

⇒ 3x – x = 10 

⇒ 2x = 10 

⇒ x = 10/2 = 5m 

∴ Width of the road = 5 m 

Now Height of the tower = √3x = 5√3 m.