Let in a right angled triangle, the smallest angle be θ. If a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then sinθ is equal to

Let $\mathrm{\angle }\mathrm{A}=\mathrm{\theta }\text{, then }\mathrm{\angle }\mathrm{B}={90}^{\circ }-\mathrm{\theta }\text{ and }{\mathrm{c}}^2={\mathrm{a}}^2+{\mathrm{b}}^2$

Since, $\mathrm{\theta }$ is smallest angle, then ‘a’ is smallest side.

$\Rightarrow \frac{1}{\mathrm{a}}$ is largest

In that case, the right angle so formed have,

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

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

Using sine law of triangle,

$\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}=\mathrm{\lambda }$

Let $\mathrm{a}=\mathrm{\lambda sin}\mathrm{A}, \mathrm{b}=\mathrm{\lambda sin}\mathrm{B}, \mathrm{c}=\mathrm{\lambda sin}\mathrm{C}$

$\begin{array}{l}\Rightarrow \mathrm{a}=\mathrm{\lambda sin}\mathrm{\theta }, \mathrm{b}=\mathrm{\lambda sin}\left({90}^{\circ }-\mathrm{\theta }\right), \mathrm{c}=\mathrm{\lambda sin}{90}^{\circ }\\ \Rightarrow \mathrm{a}=\mathrm{\lambda sin}\mathrm{\theta }, \mathrm{b}=\mathrm{\lambda cos}\mathrm{\theta }, \mathrm{c}=\mathrm{\lambda }\end{array}$

Putting in Eq. (i),

$\begin{array}{l}\frac{{\mathrm{\lambda }}^2{\cos }^2\mathrm{\theta }\cdot {\mathrm{\lambda }}^2}{{\mathrm{\lambda }}^2{\sin }^2\mathrm{\theta }}={\mathrm{\lambda }}^2{\cos }^2\mathrm{\theta }+{\mathrm{\lambda }}^2\\ \Rightarrow \frac{{\mathrm{\lambda }}^2{\cos }^2\mathrm{\theta }}{{\sin }^2\mathrm{\theta }}={\mathrm{\lambda }}^2{\cos }^2\mathrm{\theta }+{\mathrm{\lambda }}^2\\ \Rightarrow \frac{1-{\sin }^2\mathrm{\theta }}{{\sin }^2\mathrm{\theta }}={\cos }^2\mathrm{\theta }+1\end{array}$

$\begin{array}{ll}\Rightarrow & 1-{\sin }^2\mathrm{\theta }={\sin }^2\mathrm{\theta }\left(1-{\sin }^2\mathrm{\theta }+1\right)\\ \Rightarrow & {\sin }^4\mathrm{\theta }-3{\sin }^2\mathrm{\theta }+1=0\end{array}$

Let ${\sin }^2\mathrm{\theta }=\mathrm{t}$ then,

$\begin{array}{l}{\mathrm{t}}^2-3\mathrm{t}+1=0\\ \mathrm{t}=\frac{3\pm \sqrt{9-4}}{2}=\frac{3\pm \sqrt{5}}{2}\\ \therefore \mathrm{t}=\frac{3-\sqrt{5}}{2} \left[\text{ Discard }\mathrm{t}=\frac{3+\sqrt{5}}{2}\right]\end{array}$

$\Rightarrow {\sin }^2\mathrm{\theta }=\frac{3-\sqrt{5}}{2}\Rightarrow \sin \mathrm{\theta }=\frac{\sqrt{5}-1}{2}$