Question

If $\Delta$ stands for the area of a triangle ABC**,** then

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

• A. $3 \Delta$
• B. $2 \Delta$
• C. $4 \Delta$
• D. $- 4 \Delta$

Answer 1

Answer: (C)

$4 \Delta$

Answer 2

Use sine rule, $\Delta = \frac { 1 } { 2 }$ ab sin C

L.H.S.=$k ^ { 2 } \left( \sin ^ { 2 } A .2 \sin B \cos B + \sin ^ { 2 } B .2 \sin A \cos A \right)$

=$2 a b \sin C = 4 \Delta$