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Question Number 39559 by ajfour last updated on 07/Jul/18

Commented by ajfour last updated on 07/Jul/18

$F i n d α a n d r a d i u s r i n t e r m s o f a .$
	Answered by MrW3 last updated on 08/Jul/18

	$2 R \times \cos α = 1 + a . . . (i)$ $\frac{R}{\cos α} - 1 = (R + R \tan α) \sin α . . . (i i)$ $\Rightarrow R = \frac{1 + a}{2 \cos α}$ $\frac{1 + a}{2 \cos^{2} α} - 1 = \frac{1 + a}{2 \cos α} (1 + \frac{\sin α}{\cos α}) \sin α$ $\frac{1 + a - 2 \cos^{2} α}{2 \cos^{2} α} = \frac{(1 + a) (\cos α + \sin α) \sin α}{2 \cos^{2} α}$ $1 + a - 2 \cos^{2} α = (1 + a) (\sin α \cos α + \sin^{2} α)$ $2 [a - (2 \cos^{2} α - 1)] = (1 + a) (2 \sin α \cos α - (1 - {2 \sin}^{2} α) + 1)$ $2 (a - \cos 2 α) = (1 + a) (\sin 2 α - \cos 2 α + 1)$ $2 a - 2 \cos 2 α = (1 + a) \sin 2 α - (1 + a) \cos 2 α + (1 + a)$ $(a + 1) \sin 2 α - (a - 1) \cos 2 α = a - 1$ $\frac{a + 1}{\sqrt{{(a + 1)}^{2} + {(a - 1)}^{2}}} \sin 2 α - \frac{a - 1}{\sqrt{{(a + 1)}^{2} + {(a - 1)}^{2}}} \cos 2 α = \frac{a - 1}{\sqrt{{(a + 1)}^{2} + {(a - 1)}^{2}}}$ $\frac{a + 1}{\sqrt{2 (a^{2} + 1)}} \sin 2 α - \frac{a - 1}{\sqrt{2 (a^{2} + 1)}} \cos 2 α = \frac{a - 1}{\sqrt{2 (a^{2} + 1)}}$ $\cos θ \sin 2 α - \sin θ \cos 2 α = \sin θ$ $\sin (2 α - θ) = \sin θ$ $\Rightarrow 2 α - θ = θ$ $\Rightarrow α = θ = \tan^{- 1} \frac{a - 1}{a + 1}$ $\Rightarrow \cos α = \cos θ = \frac{a + 1}{\sqrt{2 (a^{2} + 1)}}$ $\Rightarrow R = \frac{1 + a}{2 \cos α} = \frac{1 + a}{2 \times \frac{a + 1}{\sqrt{2 (a^{2} + 1)}}} = \sqrt{\frac{a^{2} + 1}{2}}$
	Commented by ajfour last updated on 08/Jul/18

	$C o r r e c t S i r, t h a n k s; i t h o u g h t$ $y o u w o u l d m a k e i t s h o r t e r t h a n$ $m y s o l u t i o n; i t s a l m o s t t h e s a m e$ $l e n g t h .$
	Answered by ajfour last updated on 08/Jul/18

	$2 R \cos α = a + 1$ $R \sec α - (R + R \tan α) \sin α = 1$ $F r o m a b o v e t w o e q s .$ $\frac{1}{R} = \frac{2 \cos α}{a + 1} = \sec α - (1 + \tan α) \sin α$ $\Rightarrow \frac{2 \cos^{2} α}{a + 1} = 1 - \sin α \cos α - \sin^{2} α$ $\Rightarrow \frac{2 \cos^{2} α}{a + 1} = \cos α (\cos α - \sin α)$ $\Rightarrow \frac{2 \cos α}{\cos α - \sin α} = a + 1$ $\Rightarrow \frac{2}{1 - \tan α} = a + 1$ $\Rightarrow \tan α = 1 - \frac{2}{a + 1} = \frac{a - 1}{a + 1}$ $α = \tan^{- 1} (\frac{a - 1}{a + 1})$ $R = \frac{a + 1}{2 \cos α} = \frac{(a + 1) \sqrt{{(a - 1)}^{2} + {(a + 1)}^{2}}}{2 (a + 1)}$ $R = \sqrt{\frac{a^{2} + 1}{2}} .$
	Commented by MrW3 last updated on 08/Jul/18

	$v e r y n i c e s i r!$
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