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Question Number 68289 by Rasheed.Sindhi last updated on 08/Sep/19

A circle is divided into two equal parts

By

An arc with center on the circle .

D e t e r m i n e

(a) T h e l e n g t h o f t h e a r c

(b) T h e r a t i o i n w h i c h t h e a r c

d i v i d e s t h e d i a m e t e r

m e e t i n g t h e c e n t e r o f t h e a r c .

Answered by mr W last updated on 08/Sep/19

Commented by mr W last updated on 09/Sep/19

l e t λ = \frac{r}{R}

\cos β = \frac{r}{2 R} = \frac{λ}{2}

\cos 2 β = 2 {(\frac{λ}{2})}^{2} - 1 = \frac{λ^{2} - 2}{2}

\sin 2 β = \frac{\sqrt{2^{2} - {(λ^{2} - 2)}^{2}}}{2} = \frac{λ \sqrt{4 - λ^{2}}}{2}

α + 2 β = π

2 α = 2 π - 4 β

\sin 2 α = - \sin 4 β = - 2 \sin 2 β \cos 2 β

= - 2 \times \frac{λ \sqrt{4 - λ^{2}}}{2} \times \frac{λ^{2} - 2}{2}

= \frac{λ (2 - λ^{2}) \sqrt{4 - λ^{2}}}{2}

\frac{r^{2}}{2} (2 β - \sin 2 β) + \frac{R^{2}}{2} (2 α - \sin 2 α) = \frac{π R^{2}}{2}

\Rightarrow λ^{2} (2 \cos^{- 1} \frac{λ}{2} - \frac{λ \sqrt{4 - λ^{2}}}{2}) + 2 π - 4 \cos^{- 1} \frac{λ}{2} - \frac{λ (2 - λ^{2}) \sqrt{(2 - λ) (2 + λ)}}{2} = π

\Rightarrow 2 (2 - λ^{2}) \cos^{- 1} \frac{λ}{2} + λ \sqrt{4 - λ^{2}} = π

\Rightarrow λ \approx 1.1587

t h e d i a m e t e r i s d i v i d e d i n r a t i o

0.8413 : 1.1587

Commented by Rasheed.Sindhi last updated on 08/Sep/19

T h α n χ for this N ice A pproach S ir!

Could λ be obtained in terms of π .

Commented by mr W last updated on 08/Sep/19

i {d o n}^{'} t t h i n k t h a t t h e e q u a t i o n f o r λ

c a n b e e x a c t l y s o l v e d .

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