Question-72952

Question Number 72952 by ajfour last updated on 05/Nov/19

Commented by ajfour last updated on 05/Nov/19

A B d i v i d e s t h e c o m p o s i t e f i g u r e

o f s q u a r e m o u n t e d b y a s e m i c i r c l e

i n t w o e q u a l p a r t s . D e t e r m i n e α .

Commented by mind is power last updated on 05/Nov/19

let O center of semi circl, β angl between OB

\frac{π}{2} R^{2} + {4 R}^{2} = \frac{R^{2}}{2} . \frac{β}{} - \frac{1}{2} (R . \sin (β) . (Rcos (β) - \frac{Rsin (β)}{tg (α)})) + {(R - (Rcos β - \frac{Rsin (β)}{tg (α)})) + 2 R} . R

\Leftrightarrow \frac{π}{2} + 4 = \frac{β}{2} - \frac{1}{2} (\sin (β) \cos (β) - \frac{\sin^{2} (β)}{tg (α)}) + (3 - \cos (β) + \frac{\sin (β)}{tg (α)})

\Rightarrow x = \frac{π}{2} + 4

\Rightarrow x - \frac{β}{2} = (\frac{1}{tg (α)}) (\frac{\sin^{2} (β)}{2} + \sin (β)) - \frac{\sin (β) \cos (β)}{2} + 3 - \cos (β)

\frac{1}{tg (α)} = \frac{x - \frac{β}{2} + \frac{\sin (β) \cos (β)}{2} + \cos (β) - 3}{\frac{\sin^{2} (β)}{2} + \sin (β)}

\frac{2 R}{tg (α)} - R = R + Rcos (β) - \frac{Rsin (β)}{tg (α)}

\frac{1}{tg (α)} (2 + \sin (β)) = \frac{2 + \cos (β)}{}

\frac{1}{tg (α)} = \frac{2 + \cos (β)}{2 + \sin (β)} = \frac{x - \frac{β}{2} + \frac{\sin (β) \cos (β)}{2} + \cos (β) - 3}{\frac{\sin^{2} (β)}{2} + \sin (β)}

\Rightarrow \sin^{2} (β) + 2 \sin (β) = - 3 \sin (β) - 6 + 2 x - \frac{β}{2} (2 + \sin (β)) + \sin (β) \cos (β))

let β_{0} of this equation

\Rightarrow α = \cot^{- 1} (\frac{x - \frac{β_{0}}{2} + \frac{\sin (β_{0}) \cos (β_{0})}{2} + \cos (β_{0}) - 3}{\frac{\sin^{2} (β_{0})}{2} + \sin (β_{0})})

Commented by MJS last updated on 05/Nov/19

I found no exact solution

α \approx 54.8872 °

Commented by ajfour last updated on 05/Nov/19

T h a n k s P o w e r f u l M i n d, a n d M j S

S i r .

Answered by ajfour last updated on 05/Nov/19

Commented by ajfour last updated on 05/Nov/19

A r e a (s e c t o r B C F) = \frac{β}{2}

\tan α = \frac{2 + \sin β}{1 + \cos β} \Rightarrow \cot α = \frac{1 + \cos β}{2 + \sin β}

\frac{1}{2} (t o t a l a r e a) = A r e a (r i g h t h a l f

o f c o m p o s i t e a r e a)

\frac{π}{4} + 1 = \cot α + 2 (2 - 2 \cot α) + \frac{β}{2}

- \frac{1}{2} \times 2 (\cot α - 1) \sin β

s a y \cot α = t

\frac{π}{4} + 1 = t + 4 - 4 t + \frac{β}{2} - (t - 1) \sin β

\Rightarrow

\frac{π}{4} + 1 = 4 + \frac{β}{2} + \sin β - (3 + \sin β) t

\Rightarrow

(3 + \sin β) (\frac{1 + \cos β}{2 + \sin β}) - \frac{β}{2} - \sin β

= 3 - \frac{π}{4}

t h i s e q . y i e l d s β a n d t h e n

α = \tan^{- 1} (\frac{2 + \sin β}{1 + \cos β}) .

Answered by mr W last updated on 05/Nov/19

Commented by mr W last updated on 05/Nov/19

\tan γ = 2 \Rightarrow \sin γ = \frac{2}{\sqrt{5}} \Rightarrow \cos γ = \frac{1}{\sqrt{5}}

A C = \sqrt{5} R

Δ A C B = \frac{\sqrt{5} R^{2} \sin (π - γ + β)}{2} = \frac{\sqrt{5} R^{2} \sin (γ - β)}{2}

\frac{\sqrt{5} R^{2} \sin (γ - β)}{2} + R^{2} + \frac{(π - β) R^{2}}{2} = \frac{1}{2} (4 R^{2} + \frac{π R^{2}}{2})

\sqrt{5} \sin (γ - β) - β = 2 - \frac{π}{2}

2 \cos β - \sin β - β = 2 - \frac{π}{2}

\Rightarrow β = 0.6192

\tan α = \frac{2 R + R \sin β}{R + R \cos β} = \frac{2 + \sin β}{1 + \cos β}

\Rightarrow α = 54.888 °

Commented by ajfour last updated on 05/Nov/19

T h a n k s m r W S i r, h o p e m y a n s w e r

c o m e s t h e s a m e; {h a v e n}^{'} t c h e c k e d

y e t . .

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