Question-184609

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

Answered by ajfour last updated on 09/Jan/23

\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

N i c e! t h a n k y o u s i r .