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Question Number 71428 by TawaTawa last updated on 15/Oct/19

	Answered by ajfour last updated on 15/Oct/19

	Commented by ajfour last updated on 15/Oct/19
	$P [rcos (θ−φ), rsin (θ−φ)] tan θ=((rsin (θ−φ))/(r−a)) ⇒ φ=θ−sin^(−1) [(((r−a)tan θ)/r)] ⇒ sin φ=sin θ(√(1−[(((r−a)tan θ)/r)]^2 ))−(((r−a)sin θ)/r) Q[−rcos (φ+α), rsin (φ+α)] tan α=((rsin (φ+α))/(a+rcos (φ+α))) ⇒ asin α+rcos (φ+α)sin α =rsin (φ+α)cos α ⇒ rsin φ=asin α ⇒ r{sin θ(√(1−[(((r−a)tan θ)/r)]^2 ))−(((r−a)sin θ)/r)} = asin α .....$
	$P [r \cos (θ - ϕ), r \sin (θ - ϕ)]$ $\tan θ = \frac{r \sin (θ - ϕ)}{r - a}$ $\Rightarrow ϕ = θ - \sin^{- 1} [\frac{(r - a) \tan θ}{r}]$ $\Rightarrow \sin ϕ = \sin θ \sqrt{1 - {[\frac{(r - a) \tan θ}{r}]}^{2}} - \frac{(r - a) \sin θ}{r}$ $Q [- r \cos (ϕ + α), r \sin (ϕ + α)]$ $\tan α = \frac{r \sin (ϕ + α)}{a + r \cos (ϕ + α)}$ $\Rightarrow a \sin α + r \cos (ϕ + α) \sin α$ $= r \sin (ϕ + α) \cos α$ $\Rightarrow r \sin ϕ = a \sin α$ $\Rightarrow r {\sin θ \sqrt{1 - {[\frac{(r - a) \tan θ}{r}]}^{2}} - \frac{(r - a) \sin θ}{r}}$ $= a \sin α$ $. . . . .$
	Commented by TawaTawa last updated on 15/Oct/19

	$God bless you sir$
	Commented by ajfour last updated on 15/Oct/19

	$w h e r e t h e p u r p l e a n d b l u e a n g l e s$ $m e e t, i s t h a t m i d p o i n t o f r a d i u s ?$
	Commented by TawaTawa last updated on 15/Oct/19

	$Yes sir$
	Commented by ajfour last updated on 15/Oct/19

	$i w i s h s o m e o n e e l s e h e l p s . .$
	Commented by mind is power last updated on 15/Oct/19

	$let β = OAB$ $we have in OAQ$ $\frac{R}{\sin (α)} = \frac{a}{\sin (\emptyset)} . . 1$ $in OAB$ $\frac{a}{\sin (\emptyset)} = \frac{R}{\sin (β)} . . . 2$ $1 & 2 \Rightarrow \frac{a}{\sin (\emptyset)} = \frac{R}{\sin (β)} = \frac{R}{\sin (α)} \Rightarrow \sin (α) = \sin (β) \Rightarrow {\begin{cases} α = β \\ β = 180 - α \end{cases}$ $α = β \Rightarrow P = Q not possible \Rightarrow β = 180 - α$ $θ = PAB = 180 - β = 180 - (180 - α) = α$ $\Rightarrow α = θ$
	Commented by TawaTawa last updated on 15/Oct/19

	$God bless you sir$
	Commented by mind is power last updated on 16/Oct/19

	$y^{'} re welcom$
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