Let in a right angled triangle, the smallest angle be θ . If a...

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.

$⇒ \dfrac{1}{{a}^{2}} = \dfrac{1}{{b}^{2}} + \dfrac{1}{{c}^{2}}$

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

Use $a = 2RsinA = 2Rsin\theta$

$b = 2RsinB = 2Rsin ( 90 ° - \theta ) = 2Rcos\theta$

$c = 2RsinC = 2\sin 90 ° = 2R$

Hence, by \left(i\right)$</p><p>$\dfrac{4{R}^{2}\cos^{2}\theta \cdot 4{R}^{2}}{4{R}^{2}\sin^{2}\theta} = 4{R}^{2}\cos^{2}\theta + 4{R}^{2}$</p><p>$⇒ \cos^{2}\theta = \sin^{2}\theta\cos^{2}\theta + \sin^{2}\theta$</p><p>$⇒ 1 - \sin^{2}\theta = \sin^{2}\theta\left(1 - \sin^{2}\theta\right) + \sin^{2}\theta$</p><p>$⇒ 1 = 3\sin^{2}\theta - \sin^{4}\theta$</p><p>$⇒ \sin^{4}\theta - 3\sin^{2}\theta + 1 = 0$</p><p>$⇒ \sin^{2}\theta = \dfrac{3 - \sqrt{5}}{2} = \dfrac{5 + 1 - 2\sqrt{5}}{4} = {\left(\dfrac{\sqrt{5} - 1}{2}\right)}^{2}$</p><p>$⇒ sin\theta = \dfrac{\sqrt{5} - 1}{2}