Let in a right angled triangle, the smallest angle be $$\theta$$. If a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then $$\sin \theta$$ is equal to:

Let the right-angled triangle have sides $$a \leq b \leq c$$ with the right angle opposite the hypotenuse $$c$$, so $$c^2 = a^2 + b^2$$. The smallest angle $$\theta$$ is opposite side $$a$$, giving $$\sin\theta = \frac{a}{c}$$.

The reciprocal triangle has sides $$\frac{1}{c} \leq \frac{1}{b} \leq \frac{1}{a}$$. For this to also be a right-angled triangle, the largest side $$\frac{1}{a}$$ must be the hypotenuse: $$\frac{1}{a^2} = \frac{1}{b^2} + \frac{1}{c^2}.$$

Multiplying through by $$a^2 b^2 c^2$$: $$b^2 c^2 = a^2 c^2 + a^2 b^2$$, so $$c^2(b^2 - a^2) = a^2 b^2$$.

Substituting $$c^2 = a^2 + b^2$$: $$(a^2 + b^2)(b^2 - a^2) = a^2 b^2$$ $$b^4 - a^4 = a^2 b^2.$$

Dividing by $$a^4$$ and letting $$t = \frac{b}{a}$$: $$t^4 - t^2 - 1 = 0$$, giving $$t^2 = \frac{1 + \sqrt{5}}{2}$$ (taking the positive root).

Now, $$\sin\theta = \frac{a}{c} = \frac{1}{\sqrt{1 + t^2}} = \frac{1}{\sqrt{1 + \frac{1+\sqrt{5}}{2}}} = \frac{1}{\sqrt{\frac{3+\sqrt{5}}{2}}} = \sqrt{\frac{2}{3+\sqrt{5}}} = \sqrt{\frac{2(3-\sqrt{5})}{4}} = \sqrt{\frac{3-\sqrt{5}}{2}}.$$

Noting that $$\frac{3 - \sqrt{5}}{2} = \left(\frac{\sqrt{5}-1}{2}\right)^2$$, we get $$\sin\theta = \frac{\sqrt{5}-1}{2}$$.