Question

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

Answer 1

Let RB = x
BQR is ext of ΔPBQ
∴ PBQ = 2$\alpha \alpha = \alpha$
Now inΔ PBQ, PBQ = QPB
PQ = QB = d
Also, BRA is ext of Δ BQR
∴ QBR = 3$\alpha - 2 \alpha = \alpha$
And BRQ = 3$\alpha = 3 \alpha$ (Linear Pair)
Now in ΔBQR, by applying Sine Law, we get
d/sin (-3$\alpha$) = 3d/4 /sin$\alpha$ = x/sin²$\alpha$
d/sin 3$\alpha$ = 3d/4 /sin$\alpha$ = x/sin²$\alpha$
d/3 sin$\alpha$ 4 sin³$\alpha$ = 3d/4 sin $\alpha$ = x/2sin $\alpha$ cos $\alpha$
d/3 4sin²$\alpha$ = 3d/4 = x/2cos $\alpha$..................(I)(II)(III)
From eq. (I),I=II
d/3 4sin²$\alpha$ = 3d/4 ⇒4 = 9 12 sin²$\alpha$
sin²$\alpha$ = 5/12 ⇒cos²$\alpha$ = 7/12
Also from eq. (1) using (II) and (III) we have
3d/4 = x/2 cos$\alpha$⇒4x²² = 9 d²cos²$\alpha$
x² = 9d²/4 = 7/12 = 21/16 d²...............(3)
Again from ΔABR, we have sin 3$\alpha$ = h/x
3 sin $\alpha$ 4 sin³$\alpha$ = h/x ⇒sin$\alpha$(3 4sin²$\alpha$) = h/x
sin$\alpha$(3 4 x5/9) = h/x (using sin²$\alpha$ = 5/12)
4/3 sin$\alpha$ = h/x
Squaring both sides, we get
16/9 sin²$\alpha$ = h²/x²16/9x2/12 = h²/x²
(again using sin²$\alpha$ = 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 = 35d²
Proved
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Amit