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Clinky observes a tower PQ of height ‘h’ from a point A on the ground. She moves a distance ‘d’ towards the foot of the tower and finds that the angle of elevation has direction and finals that the angle of elevation is ‘3’ times at A. Prove that 36h2 = 35d2 .

Clinky observes a tower PQ of height ‘h’ from a point A on the ground. She moves a
distance ‘d’ towards the foot of the tower and finds that the angle of elevation has
direction and finals that the angle of elevation is ‘3’ times at A. Prove that 36h2
 = 35d2
.

Grade:

1 Answers

Latika Leekha
askIITians Faculty 165 Points
9 years ago
Hello student,
Let RB = x
In ? PBQ, angle BQR is exterior angle
∴ ∠PBQ = 2α – α = α
Now in ? PBQ, ∠ PBQ = ∠QPB
This gives PQ = QB = d
Similarly, ∠BRA is exterior angle of ? BQR
∴ ∠QBR = 3α - 2α = α
And ∠BRQ = π – 3 α (linear pair)
Now in ?BQR, by applying Sine rule, we get
d/sin (π - 3α) = 3d/4 / sin α = x/sin 2α
⇒ d/sin 3α = 3d/4 sin α = x/sin2 α
⇒ d/3 sin α – 4 sin3α = 3d/4 sin α = x/2sin α cos α
⇒ d/3 – 4 sin2 α = 3d/4 = x/2cos α . . . . . . . . . . . . . . . . . . . . . . (A)
From eq. (A), taking first two parts
⇒ d/3 – 4 sin2 α = 3d/4 ⇒ 4 = 9 – 12 sin2 α
⇒ sin2 α = 5/12 ⇒ cos2 α = 7/12
Also from eq. (A) using last two parts, we have
3d/4 = x/2 cos α ⇒ 4x2 = 9 d2 cos2 α
x2 = 9d2/4 = 7/12 = 21/16 d2 . . . . . . . . . . . . . . . . . . . . (B)
Again from ? ABR, we have sin 3α = h/x
⇒ 3 sin α – 4 sin3 α = h/x ⇒ sin α(3 – 4 sin2 α) = h/x
⇒ sin α [3 – 4 x 5/12] = h/x (using sin2 α = 5/12)
⇒ 4/3 sin α = h/x
Squaring both sides, we get
16/9 sin2 α = h2/x2 16/9 . 2/12 = h2/x2
(again using sin2 α = 5/12)
⇒ h2 = 4 . 5/9 . 3x2
⇒ h2 = 20/27 . 21/16 d2
[using value of x2 from eq. (B)]
⇒ h2 = 35/36 d2 ⇒ 36 h2 = 35 d2
Hence, 36 h2 = 35 d2.

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