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(B) Let the sides of the right-angled triangle be $a, b, c$ where $c$ is the hypotenuse. Thus,$c^2 = a^2 + b^2$.Given that the triangle formed by the

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$\frac{\sqrt{5}-1}{2}$

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(B) Let the sides of the right-angled triangle be $a, b, c$ where $c$ is the hypotenuse. Thus,$c^2 = a^2 + b^2$.Given that the triangle formed by the reciprocals of the sides is also a right-angled triangle,the largest side among $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ must be the hypotenuse. Since $c$ is the largest side,$\frac{1}{c}$ is the smallest,so $\frac{1}{a}$ is the largest.Thus,$(\frac{1}{a})^2 = (\frac{1}{b})^2 + (\frac{1}{c})^2$.Substituting $a = c \sin 	heta$ and $b = c \cos 	heta$:$\frac{1}{c^2 \sin^2 	heta} = \frac{1}{c^2 \cos^2 	heta} + \frac{1}{c^2}$$\frac{1}{\sin^2 	heta} = \frac{1}{\cos^2 	heta} + 1$$1 = \frac{\sin^2 	heta}{\cos^2 	heta} + \sin^2 	heta$$1 = 	an^2 	heta + \sin^2 	heta$$1 = \frac{\sin^2 	heta}{1 - \sin^2 	heta} + \sin^2 	heta$Let $x = \sin^2 	heta$. Then $1 = \frac{x}{1-x} + x \implies 1-x = x + x(1-x) \implies 1-x = x + x - x^2 \implies x^2 - 3x + 1 = 0$.Using the quadratic formula,$x = \frac{3 \pm \sqrt{9-4}}{2} = \frac{3 \pm \sqrt{5}}{2}$.Since $\sin^2 	heta $\sin 	heta = \sqrt{\frac{3 - \sqrt{5}}{2}} = \sqrt{\frac{6 - 2\sqrt{5}}{4}} = \frac{\sqrt{5}-1}{2}$.

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:

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