Answers

2016-02-16T21:27:33+05:30
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
Thank You.........
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Amit
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