Question-45690

Question Number 45690 by ajfour last updated on 15/Oct/18

Commented by ajfour last updated on 15/Oct/18

R a d i u s o f c i r c l e i s u n i t y .

F i n d b i n t e r m s o f a .

Answered by MrW3 last updated on 15/Oct/18

Commented by ajfour last updated on 15/Oct/18

T h a n k y o u s o m u c h S i r; i t i s

i n d e e d a m a t t e r o f f a i t h .

BEAUTIFUL!

Commented by MrW3 last updated on 15/Oct/18

t h a n k y o u t o o s i r!

Commented by MrW3 last updated on 15/Oct/18

\tan α = \frac{\sqrt{{(2 R)}^{2} - a^{2}}}{a} = \frac{\sqrt{4 R^{2} - a^{2}}}{a}

γ = 2 α

P T = R \tan γ = R \tan (2 α) = R \frac{2 \sqrt{4 R^{2} - a^{2}}}{a (1 - \frac{4 R^{2} - a^{2}}{a^{2}})} = \frac{a R \sqrt{4 R^{2} - a^{2}}}{a^{2} - 2 R^{2}}

\tan β = \frac{P T}{2 R} = \frac{a \sqrt{4 R^{2} - a^{2}}}{2 (a^{2} - 2 R^{2})}

\cos β = \frac{2 (a^{2} - 2 R^{2})}{\sqrt{a^{2} (4 R^{2} - a^{2}) + 4 {(a^{2} - 2 R^{2})}^{2}}}

\cos β = \frac{2 (a^{2} - 2 R^{2})}{\sqrt{3 a^{4} + 16 R^{4} - 12 a^{2} R^{2}}}

\cos β = \frac{2 (a^{2} - 2 R^{2})}{\sqrt{3 {(a^{2} - 2 R^{2})}^{2} + 4 R^{4}}}

b = 2 R \cos β

\Rightarrow b = \frac{4 R (a^{2} - 2 R^{2})}{\sqrt{3 {(a^{2} - 2 R^{2})}^{2} + 4 R^{4}}}

f o r R = 1 :

\Rightarrow b = \frac{4 (a^{2} - 2)}{\sqrt{3 {(a^{2} - 2)}^{2} + 4}}

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