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Question Number 176906 by HeferH last updated on 27/Sep/22

Commented by ajfour last updated on 27/Sep/22

Commented by ajfour last updated on 27/Sep/22
$Let radius of circle be r & side of square be a. centre(r, a−r) OC=(√(r^2 +(a−r)^2 )) y_B =((2(a)+3(0))/5)=((2a)/5) x_C =r+5sin α where cos α=(a/5) hence x_C =r+(√(25−a^2 )) sin x=(((((2a)/5)))/(a−r)) tan (45°−(x/2))=(r/(a−r)) 1−tan (x/2)=((r/(a−r))){1+tan (x/2)} ⇒ tan (x/2)=((1−(r/(a−r)))/(1+(r/(a−r))))=((a−2r)/a) ⇒ (((((2a)/5)))/(a−r))=((2(1−((2r)/a)))/(1+(1−((2r)/a))^2 )) say (r/a)=t ⇒ 1+(1−2t)^2 =5(1−t)(1−2t) ⇒ 2+4t^2 −4t=10t^2 −15t+5 ⇒ 6t^2 −11t+3=0 t=((11−(√(121−72)))/(12))=(1/3) sin x=(((((2a)/5)))/(a−r))=(2/(5(1−t))) hence sin x=(2/(5(1−(1/3))))=(3/5) x=sin^(−1) (3/5)=tan^(−1) (3/4)$
$L e t r a d i u s o f c i r c l e b e r & s i d e o f$ $s q u a r e b e a .$ $c e n t r e (r, a - r)$ $O C = \sqrt{r^{2} + {(a - r)}^{2}}$ $y_{B} = \frac{2 (a) + 3 (0)}{5} = \frac{2 a}{5}$ $x_{C} = r + 5 \sin α$ $w h e r e \cos α = \frac{a}{5}$ $h e n c e x_{C} = r + \sqrt{25 - a^{2}}$ $\sin x = \frac{(\frac{2 a}{5})}{a - r}$ $\tan (45 ° - \frac{x}{2}) = \frac{r}{a - r}$ $1 - \tan \frac{x}{2} = (\frac{r}{a - r}) {1 + \tan \frac{x}{2}}$ $\Rightarrow \tan \frac{x}{2} = \frac{1 - \frac{r}{a - r}}{1 + \frac{r}{a - r}} = \frac{a - 2 r}{a}$ $\Rightarrow \frac{(\frac{2 a}{5})}{a - r} = \frac{2 (1 - \frac{2 r}{a})}{1 + {(1 - \frac{2 r}{a})}^{2}}$ $s a y \frac{r}{a} = t \Rightarrow$ $1 + {(1 - 2 t)}^{2} = 5 (1 - t) (1 - 2 t)$ $\Rightarrow 2 + 4 t^{2} - 4 t = 10 t^{2} - 15 t + 5$ $\Rightarrow 6 t^{2} - 11 t + 3 = 0$ $t = \frac{11 - \sqrt{121 - 72}}{12} = \frac{1}{3}$ $\sin x = \frac{(\frac{2 a}{5})}{a - r} = \frac{2}{5 (1 - t)}$ $h e n c e \sin x = \frac{2}{5 (1 - \frac{1}{3})} = \frac{3}{5}$ $x = \sin^{- 1} \frac{3}{5} = \tan^{- 1} \frac{3}{4}$
Commented by Tawa11 last updated on 28/Sep/22

$Great sir .$
	Answered by mr W last updated on 27/Sep/22

	Commented by mr W last updated on 27/Sep/22

	$2 α + x = \frac{π}{2}$ $\Rightarrow α = \frac{π}{4} - \frac{x}{2}$ $β = \frac{x}{2}$ $r + \frac{r}{\tan α} = (2 + 3) \cos β$ $\Rightarrow r (1 + \frac{1}{\tan (\frac{π}{4} - \frac{x}{2})}) = 5 \cos \frac{x}{2} . . . (i)$ $D B = \frac{r}{\tan α} = \frac{r}{\tan (\frac{π}{4} - \frac{x}{2})}$ $\frac{D B}{\sin (\frac{π}{2} - β)} = \frac{C B}{\sin x}$ $\frac{r}{\tan (\frac{π}{4} - \frac{x}{2}) \cos \frac{x}{2}} = \frac{2}{\sin x}$ $\Rightarrow \frac{r}{\tan (\frac{π}{4} - \frac{x}{2})} = \frac{1}{\sin \frac{x}{2}} . . . (i i)$ $(i) / (i i) :$ $(1 + \frac{1}{\tan (\frac{π}{4} - \frac{x}{2})}) \tan (\frac{π}{4} - \frac{x}{2}) = 5 \cos \frac{x}{2} \sin \frac{x}{2}$ $1 + \tan (\frac{π}{4} - \frac{x}{2}) = \frac{5 \sin x}{2}$ $1 + \frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}} = \frac{5 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}$ $\frac{2}{1 + \tan \frac{x}{2}} = \frac{5 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}$ $3 \tan^{2} \frac{x}{2} + 5 \tan \frac{x}{2} - 2 = 0$ $\Rightarrow \tan \frac{x}{2} = \frac{1}{3}$ $\tan x = \frac{2 \times \frac{1}{3}}{1 - {(\frac{1}{3})}^{2}} = \frac{3}{4}$ $\Rightarrow x = \tan^{- 1} \frac{3}{4} ✓$
	Commented by Tawa11 last updated on 28/Sep/22

	$Great sir$
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