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Question Number 43265 by Cheyboy last updated on 09/Sep/18

Commented by ajfour last updated on 09/Sep/18

Commented by ajfour last updated on 09/Sep/18

$i n c l i n a t i o n o f E D w i t h h o r i z o n t a l$ $= ∠ A E D - ϕ = 90 ° - \frac{180 ° - (θ + ϕ)}{2} - ϕ$ $= \frac{θ - ϕ}{2}$ $l e t A E = A D = r$ $L e t ∠ A P G = α, ∠ A P F = β$ $A C = 2 R \sin ϕ \Rightarrow A T = \frac{A E + A C}{2}$ $\Rightarrow A T = R \sin ϕ + \frac{r}{2} = A M + T M$ $\Rightarrow T M = \frac{r}{2} = G P \sin ∠ G P M$ $\Rightarrow \frac{r}{2} = R \sin (α - ϕ) . . . . (i)$ $S i m i l a r l y o n t h e o t h e r s i d e$ $S N = \frac{r}{2} = R \sin (β - θ) . . . (i i)$ $(i) a n d (i i) \Rightarrow$ $β - α = θ - ϕ . . . (I)$ $i n c l i n a t i o n o f G F w i t h h o r i z o n t a l$ $i s = ∠ o f G P w i t h h o r i z o n t a l$ $- ∠ P G F$ $= 90 ° - α - \frac{180 - (α + β)}{2}$ $= \frac{α + β}{2} - α = \frac{β - α}{2} = \frac{θ - ϕ}{2}$ $[S e e (I)] H e n c e E D / / G F .$
Commented by Cheyboy last updated on 09/Sep/18

$Thank you sir, Really appreciated$
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