JEE Main 2021 — Properties of Triangles Question with Solution

Let $A=\theta$, then $B=90°-\theta$.

Hence, $a$ is the smallest side.

Then, in $∆ACB$

$c^2=a^2+b^2$

On taking reciprocal, longest side become shortest and vice versa.

$\Rightarrow \frac{1}{a^2}=\frac{1}{b^2}+\frac{1}{c^2}$

$\Rightarrow \frac{b^2{c}^2}{a^2}=b^2+c^2 ...\left(i\right)$

Use $a=2R\sin A=2R\sin \theta$

$b=2R\sin B=2R\sin (90°-\theta )=2R\cos \theta$

$c=2R\sin C=2\sin 90°=2R$

Hence, by $\left(i\right)$

$\frac{4R^2{\cos }^2\theta ·4R^2}{4R^2{\sin }^2\theta }=4R^2{\cos }^2\theta +4R^2$

$\Rightarrow {\cos }^2\theta ={\sin }^2\theta {\cos }^2\theta +{\sin }^2\theta$

$\Rightarrow 1-{\sin }^2\theta ={\sin }^2\theta \left(1-{\sin }^2\theta \right)+{\sin }^2\theta$

$\Rightarrow 1=3{\sin }^2\theta -{\sin }^4\theta$

$\Rightarrow {\sin }^4\theta -3{\sin }^2\theta +1=0$

$\Rightarrow {\sin }^2\theta =\frac{3-\sqrt{5}}{2}=\frac{5+1-2\sqrt{5}}{4}={\left(\frac{\sqrt{5}-1}{2}\right)}^2$

$\Rightarrow \sin \theta =\frac{\sqrt{5}-1}{2}$