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Question Number 40170 by ajfour last updated on 16/Jul/18

Commented by ajfour last updated on 16/Jul/18

Commented by ajfour last updated on 16/Jul/18

$p = \frac{1}{2 \sin α} q = \frac{1}{2 \sin β}$ $l e t s t a k e B a s o r i g i n a n d$ $x - a x i s l e f t w a r d s, y - a x i s u p w a r d s;$ $e q . o f r e d c i r c l e :$ ${(y - \frac{1}{2 \tan α})}^{2} + {(x - \frac{1}{2})}^{2} = p^{2}$ $s i m i l a r l y e q . o f b l u e c i r c l e :$ ${(y - \frac{1}{2})}^{2} + {(x - \frac{1}{2 \tan β})}^{2} = q^{2}$ $e q . o f c o m m o n c h o r d :$ $(\frac{1}{2} - \frac{1}{2 \tan α}) (2 y - \frac{1}{2} - \frac{1}{2 \tan α})$ $- (\frac{1}{2} - \frac{1}{2 \tan β}) (2 x - \frac{1}{2} - \frac{1}{2 \tan β})$ $= p^{2} - q^{2}$ $s l o p e o f c h o r d i s$ $\tan θ = \frac{\frac{1}{2} - \frac{1}{2 \tan β}}{\frac{1}{2} - \frac{1}{2 \tan α}}$ $\Rightarrow \tan θ = \frac{\tan α}{\tan β} (\frac{\tan β - 1}{\tan α - 1})$ $N o w \frac{a}{\sin θ} = \frac{1}{\sin α}$ $\Rightarrow a = \frac{\sin θ}{\sin α}$ $a = \frac{\cos α (\tan β - 1)}{\sqrt{\tan^{2} α {(\tan β - 1)}^{2} + \tan^{2} β {(\tan α - 1)}^{2}}}$ $\frac{c}{\cos θ} = \frac{1}{\sin β}$ $c = \frac{\cos β (\tan α - 1)}{\sqrt{\tan^{2} α {(\tan β - 1)}^{2} + \tan^{2} β {(\tan α - 1)}^{2}}}$ $. . . .$
Commented by MrW3 last updated on 16/Jul/18

$n i c e i l l u s t r a t i o n s i r!$ $P i s t h e i n t e r s e c t i o n o f t w o c i r c l e s,$ $t h u s w e o n l y n e e d α a n d β o r α a n d γ$ $t o d e t e r m i n e p o i n t P a n d t h e n a, b, c, d .$
	Answered by MrW3 last updated on 16/Jul/18

	Commented by MrW3 last updated on 17/Jul/18

	$\frac{1}{\sin α} = \frac{b}{\sin (α + θ)}$ $\Rightarrow b = \frac{\sin (α + θ)}{\sin α} = \cos θ + \frac{\sin θ}{\tan α}$ $\frac{1}{\sin β} = \frac{b}{\sin (β + \frac{π}{2} - θ)}$ $\Rightarrow b = \frac{\sin [\frac{π}{2} - (θ - β)]}{\sin β} = \frac{\cos (θ - β)}{\sin β} = \frac{\cos θ}{\tan β} + \sin θ$ $\cos θ + \frac{\sin θ}{\tan α} = \frac{\cos θ}{\tan β} + \sin θ$ $\tan β (1 - \tan α) \sin θ = \tan α (1 - \tan β) \cos θ$ $\Rightarrow \tan θ = \frac{\tan α (1 - \tan β)}{\tan β (1 - \tan α)} = \frac{1 - \cot β}{1 - \cot α}$ $\Rightarrow \sin θ = \frac{1 - \cot β}{\sqrt{{(1 - \cot α)}^{2} + {(1 - \cot β)}^{2}}}$ $\Rightarrow \cos θ = \frac{1 - \cot α}{\sqrt{{(1 - \cot α)}^{2} + {(1 - \cot β)}^{2}}}$ $\Rightarrow θ = \tan^{- 1} \frac{1 - \cot β}{1 - \cot α}$ $\frac{a}{\sin θ} = \frac{1}{\sin α}$ $\Rightarrow a = \frac{\sin θ}{\sin α}$ $\Rightarrow a = \frac{1 - \cot β}{\sin α \sqrt{{(1 - \cot α)}^{2} + {(1 - \cot β)}^{2}}}$ $b = \cos θ + \frac{\sin θ}{\tan α}$ $\Rightarrow b = \frac{1 - \cot α \cot β}{\sqrt{{(1 - \cot α)}^{2} + {(1 - \cot β)}^{2}}}$ $\frac{c}{\sin (\frac{π}{2} - θ)} = \frac{1}{\sin β}$ $\Rightarrow c = \frac{\cos θ}{\sin β}$ $\Rightarrow c = \frac{1 - \cot α}{\sin β \sqrt{{(1 - \cot α)}^{2} + {(1 - \cot β)}^{2}}}$ $\Rightarrow d = \frac{1 - \cot α}{\sin γ \sqrt{{(1 - \cot γ)}^{2} + {(1 - \cot α)}^{2}}}$ $b u t d c a n a l s o b e e x p r e s s e d w i t h o u t γ :$ $\Rightarrow x = b \cos θ$ $\Rightarrow y = b \sin θ$ $\Rightarrow d = \sqrt{{(1 - x)}^{2} + {(1 - y)}^{2}}$ $= \sqrt{2 + x^{2} + y^{2} - 2 (x + y)}$ $= \sqrt{2 + b^{2} - 2 b (\cos θ + \sin θ)}$ $\Rightarrow d = \sqrt{2 + \frac{(1 - \cot α \cot β) [5 - 2 (\cot α + \cot β)]}{{(1 - \cot α)}^{2} + {(1 - \cot β)}^{2}}}$ $i . e . w e o n l y n e e d t w o p a r a m e t e r s f r o m$ $α, β, γ t o d e t e r m i n e a, b, c, d .$
	Commented by ajfour last updated on 16/Jul/18

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