UPSEE Solved Model Paper 1

Let height of pole PQ = H

∠𝐵𝑃𝑄 = π/2 − 2θ
∠𝐵𝑃𝑄 = π/2 − θ
∠𝐵𝑃𝑄 = (π/2 − θ) − (π/2 − 2θ)
= θ
Now, In Δ 𝐴𝑃𝐵
∠𝐵𝐴𝑃 = ∠𝐴𝑃𝐵,
So, AB - PB = l
In Δ𝑃𝐵𝑄
PB sin 2 𝜃 = H
⇒ 𝐻 = ℓ sin 2𝜃
Alternate sol:

Let heigth of pole PQ = H, and BQ = X
In Δ𝑃𝐵𝑄,
tan 𝜃 =

H

l+x

⇒ tan 𝜃 =

H

l+( H tan2θ )

⇒ ltanθ+

Htanθ

tan2θ

= H
⇒H[1+

tanθ

( 2tanθ 1−tan2θ )

] = l tan θ
⇒ H[1−

(1−tan2θ)

2

] = l tan θ
⇒ h[1−

2ltanθ

1+tan2θ

] [

2tanθ

1+tan2θ

=sin2θ]
I. ⇒ 𝐻 = ℓ sin 2 θ