QJEEM21Mac566: JEE Main 2021 Mathematics - Class 12 JEEM21Mac566 Solution: Let in a right angled triangle, the smallest angle be . If a…

Question

Verified Solution • 2025–26 Session moderate

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 :

Accepted Answer

Let a

\Delta

ABC having C = 90

^\circ

and A =

\theta

{{\sin \theta } \over a} = {{\cos \theta } \over b} = {1 \over c}

..... (i)

Also for triangle of reciprocals

\cos A = {{{{\left( {{1 \over c}} \right)}^2} + {{\left( {{1 \over b}} \right)}^2} - {{\left( {{1 \over a}} \right)}^2}} \over {2\left( {{1 \over c}} \right)\left( {{1 \over b}} \right)}}

{1 \over {{c^2}}} + {1 \over {{{(c\cos \theta )}^2}}} = {1 \over {{{(c\sin \theta )}^2}}}

\Rightarrow 1 + {\sec ^2}\theta = \cos e{c^2}\theta

\Rightarrow {1 \over 4} = {{{{\cos }^2}\theta } \over {4{{\sin }^2}\theta {{\cos }^2}\theta }}

\Rightarrow {1 \over 4} = {{{{\cos }^2}\theta } \over {{{\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 = {{ - 4 \pm \sqrt {16 + 4} } \over 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 = {{3 - \sqrt 5 } \over 2}

\Rightarrow \sin \theta = {{\sqrt 5 - 1} \over 2}

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