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Question Number 182608 by Acem last updated on 11/Dec/22

Commented by Acem last updated on 11/Dec/22

$F i n d t h e t a n g e n t l e n g h t o f t h e t w o s m a l l c i r c l e s$ $i t i n t e r s e c t s w i t h t h e b i g c i r c l e i n t w o p o i n t s$
	Answered by mr W last updated on 11/Dec/22

	Commented by mr W last updated on 12/Dec/22

	$\cos α = \frac{7 - 2}{7 + 2} = \frac{5}{9}$ $\cos (π - β - α) = \frac{3^{2} + 9^{2} - 8^{2}}{2 \times 3 \times 9} = \frac{13}{27}$ $β = π - \cos^{- 1} \frac{13}{27} - \cos^{- 1} \frac{5}{9}$ $\cos β = - \cos (\cos^{- 1} \frac{13}{27} + \cos^{- 1} \frac{5}{9})$ $= \frac{4 \sqrt{35}}{27} \times \frac{2 \sqrt{14}}{9} - \frac{13}{27} \times \frac{5}{9} = \frac{56 \sqrt{10} - 65}{243}$ $O E = 7 + 3 \cos β$ $= 7 + 3 \times \frac{56 \sqrt{10} - 65}{243} = \frac{56 \sqrt{10} + 502}{81}$ $x = F G = 2 \times \sqrt{10^{2} - {(\frac{56 \sqrt{10} + 502}{81})}^{2}}$ $= \frac{8 \sqrt{23296 - 3514 \sqrt{10}}}{81} \approx 10.902$
	Commented by Acem last updated on 12/Dec/22

	$G r e a t S i r!$ $P l e a s e j u s t r e p l a c e α i n s t e a d γ$
	Commented by mr W last updated on 12/Dec/22

	$t h e o t h e r c h o r d :$ $\cos (α - γ) = \frac{3^{2} + 9^{2} - 8^{2}}{2 \times 3 \times 9} = \frac{13}{27}$ $γ = \cos^{- 1} \frac{5}{9} - \cos^{- 1} \frac{13}{27}$ $\cos γ = \frac{56 \sqrt{10} + 65}{243}$ ${O E}^{'} = 7 - 3 \cos γ$ $= 7 - 3 \times \frac{56 \sqrt{10} + 65}{243} = \frac{502 - 56 \sqrt{10}}{81}$ $x = 2 \times \sqrt{10^{2} - {(\frac{- 56 \sqrt{10} + 502}{81})}^{2}}$ $= \frac{8 \sqrt{23296 + 3514 \sqrt{10}}}{81} \approx 18.320$
	Commented by mr W last updated on 12/Dec/22

	Commented by Acem last updated on 12/Dec/22

	$O h w h y n o t! I w i l l n o t e i t i n m y b o o k,$ $T h a n k y o u v e r y m u c h S i r!$
	Commented by Acem last updated on 12/Dec/22

	$I f o u n d a n o t h e r w a y t o s o l v e i t, b u t {i t}^{'} s v e r y$ $a n n o y i n g a n d b o r i n g$ $\sqrt{39 - 10 y - y^{2}} + \sqrt{9 - y^{2}} = 2 \sqrt{14}$ $; y i s t h e s a m e d i s . 3 \cos β_{F o r t h e 1 s t c h o r d}$ $S o, a n d o n h e r e u s i n g c o s i n e l a w s i s b e t t e r$ $t h a n P y t h a g o r a s o n e$ $H e r e c o s i n e l a w s a r e a s t h e m a g i c a l s o l u t i o n!$
	Answered by Acem last updated on 12/Dec/22

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