Question

The angle of elevation of the top of a tower as observed from a point on the ground isα and on moving 'a' metre towards the tower the angle of elevation isβ.Prove that the height of the tower is

a tanα.tanβ

tanβ- tanα

Answer 1

Given :

The angle of elevation of the top of a tower as observed from a point on the ground isα and on moving 'a' metre towards the tower the angle of elevation isβ.

To prove :

h = a[(tanαtanβ)/(tanβ- tanα)]

Solution :

From the figure ,

Let the height of the tower = AB = h

<BDA = α

<BCA = β

DC = a

Let CA = x

i ) In ∆BDC , <BAD = 90°

tanα = BA/DC

=> tanα = h/(a+x)

Do cross multiplication, we get

=> (a+x) = h/tanα

=> x = h/tanα - a -----(1)

ii) In ∆BCA , <BAC = 90°

tanβ = BA/CA

=> tanβ = h/x

=> x = h/tanβ -----(2)

(1) = (2)

=>( h/tanα) - a = h/tanβ

=> h/tanα - h/tanβ = a

=> h[ 1/tanα - 1/tanβ] = a

=> h[(tanβ - tanα)/(tanαtanβ)]=a

=> h = a[(tanαtanβ)/(tanβ- tanα)]

Therefore,

Height of the tower

(h) = a[(tanαtanβ)/(tanβ- tanα)]

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Answer 2

hope you understand ,

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