JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Properties of Triangle

Let the sides be $a<b<c$, so $c$ is the hypotenuse, $a^2+b^2=c^2$, and $\theta$ is the angle opposite $a$. Among the reciprocals $\frac{1}{a}>\frac{1}{b}>\frac{1}{c}$, so $\frac{1}{a}$ is the hypotenuse of the second right triangle:

Clearing denominators gives $a^2\left(b^2+c^2\right)=b^2{c}^2$, and substituting $b^2=c^2-a^2$,

Dividing by $c^4$ and writing $s=\frac{a^2}{c^2}={\sin }^2\theta$ gives $s^2-3s+1=0$, whose roots are $\frac{3\pm \sqrt{5}}{2}$. Only $s=\frac{3-\sqrt{5}}{2}$ is less than $1$, and

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