Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
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
$let O center of semi circl,β angl between OB (π/2)R^2 +4R^2 =(R^2 /2).(β/)−(1/2)(R.sin(β).(Rcos(β)−((Rsin(β))/(tg(α)))))+{(R−(Rcosβ−((Rsin(β))/(tg(α)))))+2R}.R ⇔(π/2)+4=(β/2) −(1/2)(sin(β)cos(β)−((sin^2 (β))/(tg(α))))+(3−cos(β)+((sin(β))/(tg(α)))) ⇒x=(π/2)+4 ⇒x−(β/2)=((1/(tg(α))))(((sin^2 (β))/2)+sin(β))−((sin(β)cos(β))/2)+3−cos(β) (1/(tg(α)))=((x−(β/2)+((sin(β)cos(β))/2)+cos(β)−3)/(((sin^2 (β))/2)+sin(β))) ((2R)/(tg(α)))−R=R+Rcos(β)−((Rsin(β))/(tg(α))) (1/(tg(α)))(2+sin(β))=((2+cos(β))/) (1/(tg(α)))=((2+cos(β))/(2+sin(β)))=((x−(β/2)+((sin(β)cos(β))/2)+cos(β)−3)/(((sin^2 (β))/2)+sin(β))) ⇒sin^2 (β)+2sin(β)=−3sin(β)−6+2x−(β/2)(2+sin(β))+sin(β)cos(β)) let β_0 of this equation ⇒α=cot^(−1) (((x−(β_0 /2)+((sin(β_0 )cos(β_0 ))/2)+cos(β_0 )−3)/(((sin^2 (β_0 ))/2)+sin(β_0 ))))$
$let O center of semi circl, β angl between OB$ $\frac{π}{2} R^{2} + {4 R}^{2} = \frac{R^{2}}{2} . \frac{β}{} - \frac{1}{2} (R . \sin (β) . (Rcos (β) - \frac{Rsin (β)}{tg (α)})) + {(R - (Rcos β - \frac{Rsin (β)}{tg (α)})) + 2 R} . R$ $\Leftrightarrow \frac{π}{2} + 4 = \frac{β}{2} - \frac{1}{2} (\sin (β) \cos (β) - \frac{\sin^{2} (β)}{tg (α)}) + (3 - \cos (β) + \frac{\sin (β)}{tg (α)})$ $\Rightarrow x = \frac{π}{2} + 4$ $\Rightarrow x - \frac{β}{2} = (\frac{1}{tg (α)}) (\frac{\sin^{2} (β)}{2} + \sin (β)) - \frac{\sin (β) \cos (β)}{2} + 3 - \cos (β)$ $\frac{1}{tg (α)} = \frac{x - \frac{β}{2} + \frac{\sin (β) \cos (β)}{2} + \cos (β) - 3}{\frac{\sin^{2} (β)}{2} + \sin (β)}$ $\frac{2 R}{tg (α)} - R = R + Rcos (β) - \frac{Rsin (β)}{tg (α)}$ $\frac{1}{tg (α)} (2 + \sin (β)) = \frac{2 + \cos (β)}{}$ $\frac{1}{tg (α)} = \frac{2 + \cos (β)}{2 + \sin (β)} = \frac{x - \frac{β}{2} + \frac{\sin (β) \cos (β)}{2} + \cos (β) - 3}{\frac{\sin^{2} (β)}{2} + \sin (β)}$ $\Rightarrow \sin^{2} (β) + 2 \sin (β) = - 3 \sin (β) - 6 + 2 x - \frac{β}{2} (2 + \sin (β)) + \sin (β) \cos (β))$ $let β_{0} of this equation$ $\Rightarrow α = \cot^{- 1} (\frac{x - \frac{β_{0}}{2} + \frac{\sin (β_{0}) \cos (β_{0})}{2} + \cos (β_{0}) - 3}{\frac{\sin^{2} (β_{0})}{2} + \sin (β_{0})})$
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

	$T h a n k s m r W S i r, h o p e m y a n s w e r$ $c o m e s t h e s a m e; {h a v e n}^{'} t c h e c k e d$ $y e t . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum