Question

A TV tower stands vertically on a bank of a Canal from a point on the other bank directly opposite the tower the angle of elevation of the top of the tower is 60° from. 20 m away from this point on the same bank the angle of elevation of the top of the tower is 30° find the height of the tower and the width of the canal...​

Answer 1

Solutions :

let

PQ = h meter be the height of the tower

BQ = x metre be the width of the canal

angle PQB = 60°

Now the angle of elevation of the top of of tower from the point A = 30°

angle PAQ = 30° where AB = 20 meter

from ΔPBQ ,

h/x = tan60°

h = x√3. ........(1)

from ΔPAQ ,

h/20 + x = tan30° = 1/√3

h = 20 + x/√3. .........(2)

from 1 and 2 we have

x = 10 m

from (1),

h = 10√3 m

=> hence the height of the tower is 10√3 m and width of the canal is 10 m

Answer 2

Consider a ∆ ABD, having AB = height of T.V. tower.

$\angle{ACB}$ = 60° and $\angle{ADB}$ = 30°

DC = 20 m

• Let BC = x (width of river)

In ∆ABC

→ tan60° = [tex]\dfrac{AB}{BC} [/tex]

→ √3 = [tex]\dfrac{h}{x} [/tex]

By Cross-multiplication

→ h = x√3 ......(1)

___________________________

In ∆ABD

→ tan30° = [tex]\dfrac{AB}{BD} [/tex]

→ [tex]\dfrac{1}{\sqrt{3}} [/tex]

= [tex]\dfrac{AB}{BC\:+\:CD} [/tex]

→ [tex]\dfrac{1}{\sqrt{3}} [/tex]

= [tex]\dfrac{h}{x\:+\:20} [/tex]

By Cross-multiplication

→ h√3 = x + 20

→ (x√3)√3 = x + 20 [From 1]

→ 3x = x + 20

→ 3x - x = 20

→ 2x = 20

→ x = 10 m

Put value of x in (1)

→ h = 10√3 m

____________________________

Height of tower = 10√3 m and width of river = 10 m