Question

The angle of elevation of the top of building at a point on the level ground is 45degree . After moving 200m towards the building along the same horizontal line, the angle of elevation of the building is 60degree . Find the height of the building

Answer 1

Question: The angle of elevation of a tower at a point is 45∘. After going 40 m towards the foot of the tower, the angle of elevation of the tower becomes 60∘. Find the height of the tower.

Step-by-step explanation:

GIVEN:

∠ADB = 60° , ∠ACB = 45° , CD= 40 m, BD= x m  

Let AB = h m be the height of the tower.

In ∆ ABD

tan 60° = AB/BD = P /B

√3 = h /x

h = x √3

x = h/√3 ………………………... (1)

In ∆ ABC

tan 45° = AB/BC

tan 45° = AB/(BD+DC)

1 = AB/(x+40)

1= h /(x+40)

h = x + 40 ………………………. (2)

Put the value of x from eq 1 in eq 2.

h = x + 40

h = h/√3 +40

h - h/ √3  = 40

(√3h -h ) /√3 = 40

h (√3 -1) = 40√3

h = 40√3 /(√3 -1)

h = [40√3 /(√3 - 1)] x [(√3 +1)/(√3 +1)]

[By rationalising]

h = [120 + 40√3] / [(√3)² - 1²]

h = [120 + 40× 1.732] / 2

h  = (120 + 69.28)/2 = 189.28/2 = 94.64 m

Hence, the height of the tower is 94.64 m.