Question

the angle of elevation theta of a vertical tower from a point on the ground is such that its tangent is (5/12).on walking 192 meters towards in the same straight line , the tangent of the angle of elevation theta a is found to be (3/4) . find the height of the tower

Answer 1

Answer:

The height of the pole will be 180 meters.

Step-by-step explanation:

let the height of the tower be h

and the distance of the base from the first point be b

At the first point,

tanФ = 5/12

=> h/b = 5/12

=> b = 12h/5

After walking 192 meters towards the pole, tangent becomes 3/4

=> tanФ₁ = 3/4

=> h/(b-192) = 3/4

=> 4h = 3(b-192)

=> 4h = 3(12h/5 - 192)

=> 4h = 36h/5 - 576

=> 36h/5 - 4h = 576

=> 16h/5 = 576

=> h = 576 x 5/16 = 180

Hence the height of the pole will be 180 meters.

Answer 2

Solution:

Let the height of the tower be h- meter.

In triangle ACD

∴ tan ∠ADC = h/x

⇔ 3/4 = h/x

therefor x = 4h/3 ---- (1)

now in triangle ABC

tan∠ABC = h/x+192

⇔5/12 = h/x+192 (2)

now putting the value of x  from equation 1 in eq.2 we get

5/12 = h/4h/3+192

5/12 = 3h/576+4h

20h +2880 =36h

⇔ 36h-20h = 2880

⇒16h = 2880

⇒ h = 2880/16 = 180

hence the height of the tower = 180m