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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)))

$$\mathrm{cos}\:\left(\mathrm{2}\alpha−\mathrm{90}°\right)=\frac{\mathrm{2}}{{R}} \\ $$$$\Rightarrow\:\:\mathrm{sin}\:\mathrm{2}\alpha=\frac{\mathrm{2}}{{R}}={t} \\ $$$$\mathrm{tan}\:\alpha=\frac{{R}+\sqrt{{R}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{1}+\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{\mathrm{2}{t}} \\ $$$$\mathrm{sin}\:\mathrm{2}\alpha=\frac{\mathrm{2tan}\:\alpha}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \alpha} \\ $$$$\Rightarrow \\ $$$$\:{t}^{\mathrm{2}} \left\{\mathrm{1}+\frac{\mathrm{2}−{t}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{\mathrm{4}{t}^{\mathrm{2}} }\right\}=\mathrm{1}+\sqrt{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\mathrm{3}{t}^{\mathrm{2}} +\mathrm{2}+\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }=\mathrm{4}+\mathrm{4}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\:\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} +}\mathrm{3}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)−\mathrm{1}=\mathrm{0} \\ $$$${hence}\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{3}}+\sqrt{\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\:=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\:\:\:\:{t}=\pm\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$$\mathrm{tan}\:\alpha=\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)}{\left(\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{3}}\right)}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$ \\ $$

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

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

R=(a/(2 sin A)) R=((√(10))/(2 sin 45°))=(√5) tan α=((R+(√(R^2 −2^2 )))/2)=(((√5)+1)/2)

$${R}=\frac{{a}}{\mathrm{2}\:\mathrm{sin}\:{A}} \\ $$$${R}=\frac{\sqrt{\mathrm{10}}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{45}°}=\sqrt{\mathrm{5}} \\ $$$$\mathrm{tan}\:\alpha=\frac{{R}+\sqrt{{R}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }}{\mathrm{2}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$

Commented by cherokeesay last updated on 09/Jan/23

Nice ! thank you sir.

$${Nice}\:!\:{thank}\:{you}\:{sir}. \\ $$

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