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Question Number 161367 by amin96 last updated on 17/Dec/21

Commented by 1549442205PVT last updated on 17/Dec/21

$I t h i n k t h a t t h e c o n d i t i o n o f t h e p r o b e m {i s n}^{'} t c l e a r$ $t h e r a d i u s o f t h e a r c w h i c h l i m i t s s m a l l$ $c i r c l e {i s n}^{'} t g i v e n y e t$
	Answered by mr W last updated on 17/Dec/21

	Commented by mr W last updated on 17/Dec/21

	$B C = R - r$ $B E = \sqrt{{(R - r)}^{2} - r^{2}} = \sqrt{R^{2} - 2 R r}$ $O B = R$ $O E = R + \sqrt{R^{2} - 2 R r}$ $O C = \sqrt{r^{2} + {(R + \sqrt{R^{2} - 2 R r})}^{2}} = \sqrt{2 R^{2} - 2 R r + r^{2} + 2 R \sqrt{R^{2} - 2 R r}}$ $\cos α = \frac{O E}{O C}, \sin α = \frac{E C}{O C}$ $β = \frac{π}{3} - α$ $A C = R + r$ ${A C}^{2} = {O A}^{2} + {O C}^{2} - 2 \times O A \times O C \times \cos (\frac{π}{3} - α)$ ${A C}^{2} = {O A}^{2} + {O C}^{2} - O A \times O C (\cos α + \sqrt{3} \sin α)$ ${A C}^{2} = {O A}^{2} + {O C}^{2} - O A (O E + \sqrt{3} E C)$ ${(R + r)}^{2} = R^{2} + 2 R^{2} - 2 R r + r^{2} + 2 R \sqrt{R^{2} - 2 R r} - R (R + \sqrt{R^{2} - 2 R r} + \sqrt{3} r)$ $\sqrt{R^{2} - 2 R r} = (4 + \sqrt{3}) r - R$ ${(4 + \sqrt{3})}^{2} r = 2 (3 + \sqrt{3}) R$ $\Rightarrow \frac{r}{R} = \frac{2 (3 + \sqrt{3})}{{(4 + \sqrt{3})}^{2}} = \frac{2 (33 - 5 \sqrt{3})}{169} \approx 0.288044$
	Commented by Tawa11 last updated on 18/Dec/21

	$Great sir$
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