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Question Number 165217 by ajfour last updated on 27/Jan/22

Commented by mr W last updated on 28/Jan/22

$i t s h o u l d b e p o s s i b l e f o r a n y a > 0 .$
Commented by mr W last updated on 28/Jan/22

$n i c e q u e s t i o n s i r!$ $i t s e e m s t o h a v e n o u n i q u e s o l u t i o n .$ $t h a t m e a n s i t i s p o s s i b l e f o r a r a n g e$ $o f v a l u e s o f a, n o t a s i n g l e o n e .$
Commented by mr W last updated on 28/Jan/22

Commented by mr W last updated on 28/Jan/22

Commented by ajfour last updated on 28/Jan/22

$L e t s h a v e t h e r a n g e t h e n . .$ $T h a n k s S i r!$
	Answered by mr W last updated on 28/Jan/22

	Commented by mr W last updated on 28/Jan/22

	$w e t a k e O a s o r i g i n .$ $l e t m = \tan θ, k = \tan φ$ $A (0, 1)$ $B (a, 0)$ $C_{1} (a - r, 1 - r)$ $C_{2} (h, r)$ $e q n . o f O E :$ $m x - y = 0$ $e q n . o f A D :$ $k x + y - 1 = 0$ $r \sqrt{m^{2} + 1} = (a - r) m - (1 - r)$ $r \sqrt{m^{2} + 1} = h m - r$ $(a - r) m - (1 - r) = h m - r$ $\Rightarrow h = a - r - \frac{1 - 2 r}{m} ✓$ $r \sqrt{m^{2} + 1} = (a - r) m - (1 - r)$ $\Rightarrow r = \frac{a m - 1}{m + \sqrt{m^{2} - 1} - 1} ✓$ $r \sqrt{k^{2} + 1} = (a - r) k + (1 - r) - 1$ $r \sqrt{k^{2} + 1} = - (h k + r - 1)$ $(a - r) k + (1 - r) - 1 = - (h k + r - 1)$ $1 - (a - r) k = h k$ $\Rightarrow h = \frac{1}{k} - (a - r) ✓$ $r \sqrt{k^{2} + 1} = - [1 - (a - r) k + r - 1]$ $\Rightarrow r = \frac{a k}{1 + k + \sqrt{k^{2} + 1}} ✓$ $\frac{1}{k} - a + r = a - r - \frac{1 - 2 r}{m}$ $\frac{1}{k} = 2 (a - r) - \frac{1 - 2 r}{m}$ $\frac{1}{k} = 2 a - \frac{1}{m} + 2 (\frac{1}{m} - 1) r$ $\Rightarrow k = \frac{1}{2 a - \frac{1}{m} + 2 (\frac{1}{m} - 1) r} ✓$ $f o r a n y g i v e n a w e c a n o b t a i n m f r o m$ $\frac{a m - 1}{m + \sqrt{m^{2} - 1} - 1} = \frac{a k}{1 + k + \sqrt{k^{2} + 1}}$
	Commented by mr W last updated on 28/Jan/22

	Commented by mr W last updated on 28/Jan/22

	Commented by mr W last updated on 28/Jan/22

	Commented by ajfour last updated on 28/Jan/22

	$T h a n k s s i r, i w i l l g o t h r o u g h . .$
	Commented by Tawa11 last updated on 28/Jan/22

	$Great sir$
	Answered by ajfour last updated on 28/Jan/22

	Commented by ajfour last updated on 28/Jan/22

	$\tan ϕ = \frac{r}{b + r}; \tan θ = \frac{r}{a - r}$ $\cos 2 θ = r + r cosec ϕ \sin (2 θ + ϕ)$ $(a + b) (1) = \cot 2 ϕ + 2 r cosec 2 ϕ$ $. . . . .$
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