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Question Number 39666 by ajfour last updated on 09/Jul/18

Commented by ajfour last updated on 09/Jul/18

$I n t e r m s o f a a n d θ, f i n d$ $x, y, α, β, a n d r . (y = A D)$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18

	$4 θ = α + β$ $s i n α = \frac{1}{r} s i n β = \frac{a}{r}$ $c o s (α + β) = \frac{2 r^{2} - {(a + 1)}^{2}}{2 r^{2}}$ $2 r^{2} c o s (4 θ) - 2 r^{2} = - {(a + 1)}^{2}$ $2 r^{2} (1 - c o s 4 θ) = {(a + 1)}^{2}$ $2 r^{2} \times 2 {s i n}^{2} 2 θ = {(a + 1)}^{2}$ $4 r^{2} {s i n}^{2} 2 θ = {(a + 1)}^{2}$ $2 r s i n 2 θ = (a + 1)$ $r = \frac{a + 1}{2 s i n 2 θ}$ $s i n α = \frac{1}{r} = \frac{2 s i n 2 θ}{a + 1}$ $α = {s i n}^{- 1} (\frac{2 s i n 2 θ}{a + 1})$ $s i n β = \frac{a}{r} = \frac{a}{a + 1} \times 2 s i n 2 (θ)$ $β . = {s i n}^{- 1} (\frac{a}{a + 1} . 2 s i n (2 θ)}$ $\frac{x}{y} = \frac{1}{a} s o \frac{x}{1} = \frac{y}{a} k x = k a n d y = a k$ $c o s 2 θ = \frac{x^{2} + y^{2} - {(a + 1)}^{2}}{2 x y}$ $c o s 2 θ = \frac{k^{2} + a^{2} k^{2} - {(a + 1)}^{2}}{2 {a k}^{2}}$ $k^{2} + a^{2} k^{2} - 2 {a k}^{2} c o s 2 θ = {(a + 1)}^{2}$ $k^{2} (1 + a^{2} - 2 a c o s 2 θ) = {(a + 1)}^{2}$ $k = \frac{a + 1}{\sqrt{1 + a^{2} - 2 a c o s 2 θ}}$ $x = \frac{a + 1}{\sqrt{1 + a^{2} - 2 a c o s 2 θ}} y = a \times \frac{a + 1}{\sqrt{1 + a^{2} - 2 a c o s 2 θ}}$
	Commented by ajfour last updated on 10/Jul/18

	$E x c e l l e n t, T a n m a y S i r . T h a n k y o u .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18

	$q u a l i t y o f q u e s t i o n s y o u p o s t r e a l l y e y e o p e n e r$ $a n d a c t i v a t e s l e e p i n g c e l l s o f b r a i n t o g e t u p$ $a n d d o w o r k . . .$
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