If the area of a triangle $ABC$ is $\Delta $, then ${a^2}\sin 2B + {b^2}\sin 2A$ is equal to

JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

1 Easy

1 of 1

Application +4 | -1

If the area of a triangle

ABC

\Delta

, then

{a^2}\sin 2B + {b^2}\sin 2A

is equal to

View answer and explanation

Correct option: C

4\Delta

Option C (

4\Delta

) is correct because it is derived from the area of a triangle and trigonometric identities.

Practice All Chapter MCQs →

{p_1},{p_2},{p_3}

\Delta \;ABC

A,B,C

\Delta

p_1^{ - 2} + p_2^{ - 2} + p_3^{ - 2}

5k,\;6k,\;5k

k

\Delta \,ABC

a = \left( {b - c} \right)\sec \theta

\frac{{2\,\sqrt {bc} }}{{b - c}}\,\,\sin \,\,\frac{A}{2} =

\Delta \;ABC,\frac{{b + c}}{a}

\Delta = {a^2} - {\left( {b - c} \right)^2}

\Delta

\Delta \;ABC

\tan A

A

2 s