If the area of a $\triangle A B C$ be $\lambda$, then $a^2 \sin 2 B+b^2 \sin 2 A$ is equal to

JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

1 Easy

1 of 1

Application +4 | -1

If the area of a

\triangle A B C

\lambda

, then

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

is equal to

View answer and explanation

Correct option: C

4 \lambda

Option C (

4 \lambda

) is correct because using trigonometric identities and the sine rule, the expression simplifies to four times the area of the triangle.

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{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