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Question Number 184609 by cherokeesay last updated on 09/Jan/23

	Answered by ajfour last updated on 09/Jan/23
	$cos (2α−90°)=(2/R) ⇒ sin 2α=(2/R)=t tan α=((R+(√(R^2 −4)))/2)=((1+(√(1−t^2 )))/(2t)) sin 2α=((2tan α)/(1+tan^2 α)) ⇒ t^2 {1+((2−t^2 +2(√(1−t^2 )))/(4t^2 ))}=1+(√(1−t^2 )) ⇒ 3t^2 +2+2(√(1−t^2 ))=4+4(√(1−t^2 )) ⇒ 2(√(1−t^2 +))3(1−t^2 )−1=0 hence (√(1−t^2 ))=−(1/3)+(√((1/9)+(1/3))) (√(1−t^2 )) =(1/3) ⇒ t=±((2(√2))/3) tan α=(((1+(1/3)))/((((4(√2))/3))))=(1/( (√2)))$
	$\cos (2 α - 90 °) = \frac{2}{R}$ $\Rightarrow \sin 2 α = \frac{2}{R} = t$ $\tan α = \frac{R + \sqrt{R^{2} - 4}}{2} = \frac{1 + \sqrt{1 - t^{2}}}{2 t}$ $\sin 2 α = \frac{2 \tan α}{1 + \tan^{2} α}$ $\Rightarrow$ $t^{2} {1 + \frac{2 - t^{2} + 2 \sqrt{1 - t^{2}}}{4 t^{2}}} = 1 + \sqrt{1 - t^{2}}$ $\Rightarrow 3 t^{2} + 2 + 2 \sqrt{1 - t^{2}} = 4 + 4 \sqrt{1 - t^{2}}$ $\Rightarrow 2 \sqrt{1 - t^{2} +} 3 (1 - t^{2}) - 1 = 0$ $h e n c e \sqrt{1 - t^{2}} = - \frac{1}{3} + \sqrt{\frac{1}{9} + \frac{1}{3}}$ $\sqrt{1 - t^{2}} = \frac{1}{3}$ $\Rightarrow t = \pm \frac{2 \sqrt{2}}{3}$ $\tan α = \frac{(1 + \frac{1}{3})}{(\frac{4 \sqrt{2}}{3})} = \frac{1}{\sqrt{2}}$
	Answered by mr W last updated on 09/Jan/23

	Commented by mr W last updated on 09/Jan/23

	$R = \frac{a}{2 \sin A}$ $R = \frac{\sqrt{10}}{2 \sin 45 °} = \sqrt{5}$ $\tan α = \frac{R + \sqrt{R^{2} - 2^{2}}}{2} = \frac{\sqrt{5} + 1}{2}$
	Commented by cherokeesay last updated on 09/Jan/23

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