Question

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 :

• A. $\frac{\sqrt{5}+1}{4}$
• B. $\frac{\sqrt{2}-1}{2}$
• C. $\frac{\sqrt{5}-1}{2}$
• D. $\frac{\sqrt{5}-1}{4}$

Answer 1

The correct option is C $\frac{\sqrt{5}-1}{2}$

Let a $\Delta ABC$ having $C={90}^{\circ }\text{and}A=\theta$
$\frac{\sin \theta }{a}=\frac{\cos \theta }{b}=\frac{1}{c}...\left(i\right)$
Also for triangle of reciprocals
$\cos A^{\prime }=\frac{{\left(\frac{1}{c}\right)}^2+{\left(\frac{1}{b}\right)}^2-{\left(\frac{1}{a}\right)}^2}{2\left(\frac{1}{c}\right)\left(\frac{1}{b}\right)}$ and $A^{\prime }={90}^0$

$\Rightarrow \frac{1}{c^2}+\frac{1}{(ccos\theta {)}^2}=\frac{1}{(c\sin \theta {)}^2}$

$\Rightarrow 1+{\sec }^2\theta =cose{c}^2\theta$
$\Rightarrow \frac{1}{4}=\frac{\cos 2\theta }{4{\sin }^2\theta {\cos }^2\theta }$
$\Rightarrow \frac{1}{4}=\frac{\cos 2\theta }{{\sin }^{2}2\theta }$
$\Rightarrow 1-{\cos }^{2}2\theta =4\cos 2\theta$
${\cos }^{2}2\theta +4\cos 2\theta -1=0$
$\cos 2\theta =\frac{-4\pm \sqrt{16+4}}{2}$
$\cos 2\theta =-2\pm \sqrt{5}$
$\cos 2\theta =\sqrt{5}-2=1-2{\sin }^2\theta$
$\Rightarrow 2{\sin }^2\theta =3-\sqrt{5}$
$\Rightarrow {\sin }^2\theta =\frac{3-\sqrt{5}}{2}=\frac{6-2\sqrt{5}}{4}$
$\Rightarrow \sin \theta =\frac{\sqrt{5}-1}{2}$