QJEEM21Ma48a6: JEE Main 2021 Mathematics - Class 11 JEEM21Ma48a6 Solution: If in a triangle ABC, AB = 5 units, and radius of circumcircle…

Question

Verified Solution • 2025–26 Session moderate

If in a triangle ABC, AB = 5 units,

\angle B = {\cos ^{ - 1}}\left( {{3 \over 5}} \right)

and radius of circumcircle of

\Delta

ABC is 5 units, then the area (in sq. units) of

\Delta

ABC is :

Accepted Answer

As,

\cos B = {3 \over 5} \Rightarrow B = 53^\circ

As,

R = 5 \Rightarrow {c \over {\sin c}} = 2R

\Rightarrow {5 \over {10}} = \sin c \Rightarrow C = 30^\circ

Now,

{b \over {\sin B}} = 2R \Rightarrow b = 2(5)\left( {{4 \over 5}} \right) = 8

Now, by cosine formula

\cos B = {{{a^2} + {c^2} - {b^2}} \over {2ac}}

\Rightarrow {3 \over 5} = {{{a^2} + 25 - 64} \over {2(5)a}}

\Rightarrow {a^2} - 6a - 3g = 0

\therefore

a = {{6 \pm \sqrt {192} } \over 2} = {{6 \pm 8\sqrt 3 } \over 2}

\Rightarrow 3 + 4\sqrt 3

(Reject

a = 3 - 4\sqrt 3

)

Now,

\Delta = {{abc} \over {4R}} = {{(3 + 4\sqrt 3 )(8)(5)} \over {4(5)}} = 2(3 + 4\sqrt 3 )

\Rightarrow \Delta = (6 + 8\sqrt 3 )

\Rightarrow

Option (3) is correct.

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