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# A man observes a tower AB of height h From a point P on the ground. He moves a distance d towards the foot of the tower and finds that the angle of elevation has doubled. He further moves a distance 3/4 d in the same direction and finds that the angle of elevation is three times that at P. Prove that 36h2 = 35d2.

Jitender Pal
7 years ago
Hello Student,
Let RB = x
∠BQR is ext ∠ of ∆ PBQ
∴ ∠PBQ = 2 α – α = α
Now in ∆ PBQ, ∠ PBQ, = ∠QPB
⇒ PQ = QB = d
Also ∠BRA is ext. ∠ of ∆ BQR
∴ ∠QBR = 3α - 2α = α
And ∠BRQ = π – 3 α (linear pair)
Now in ∆BQR, by applying Sine Law, we get
d/sin (π - 3α) = 3d/4 / sin α = x/sin 2 α
⇒ d/sin 3 α = 3d/4 sin α = x/sin2 α
⇒ d/3 sin α – 4 sin3 α = 3d/4 sin α = x/2sin α cos α
⇒ d/3 – 4 sin2 α = 3d/4 = x/2cos α . . . . . . . . . . . . . . . . . . . . . . (I)
1. (II) (III)
From eq. (I), I = II
⇒ d/3 – 4 sin2 α = 3d/4 ⇒ 4 = 9 – 12 sin2 α
⇒ sin2 α = 5/12 ⇒ cos2 α = 7/12
Also from eq. (I) using (II) and (III), we have
3d/4 = x/2 cos α ⇒ 4x2 = 9 d2 cos2 α
x2 = 9d2/4 = 7/12 = 21/16 d2 . . . . . . . . . . . . . . . . . . . . (3)
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 x 2/12 = h2/x2
(again using sin2 α = 5/12)
⇒ h2 = 4 x 5/9 x 3 x2 ⇒ h2 = 20/27 x 21/16 d2
[using value of x2 from eq. (3)]
⇒ h2 = 35/36 d2 ⇒ 36 h2 = 35 d2
Which was to be proved.

Thanks
Jitender Pal