Question

A person observed the angle of elevation of the top of a tower as 30°. He walked 50 m towards the foot of the tower along level ground and found the angle of elevation of the top of the tower as 60°. Find the height of the tower.

Answer 1

Answer:

The height of the tower is 43.25 m.

Step-by-step explanation:

Let AB be the height of the tower (h).

Given :  

Angle of elevation of the top of the tower (θ), ∠ADB = 30°

He walks CD = 50 m towards the foot of the tower,then the Angle of elevation of  the top of the tower ,(θ), ∠ACB = 60°

Let CB = x m

In right angle triangle, ∆ABC,

tan θ  = P/ B

tan 60° = AB/BC

√3 = h/x

x = h/√3 ……...…..(1)

In right angle triangle, ∆ABD

tan θ  = P/ B

tan 30° = AB/BD

1/√3 = AB/(BC+CD)

1/√3 = h/(x + 50)

√3h = x + 50

√3h = h/√3 + 50

[From eq 1]

√3h - h/√3 = 50

(√3 ×√3 h - h)/√3 = 50

(3h - h)/√3 = 50

2h = 50√3

h = (50√3)/2

h = 25√3

h = 25 × 1.73

h = 43.25 m

Hence, the height of the tower is 43.25 m

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Step-by-step explanation:

HEIGHT OF TOWER

tan 30 = h/50+x

50+x=h√3

50=h√3-x

tan 60 = h/x

x√3=h

50=3x-x

50=2x

x=25

h=x√3

hence height of tower is 25√3

43.3072 is the answer...

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43.3072 IS ANSWER