Question

In a triangle $ABC,\frac{b^2\sin 2C+c^2\sin 2B}{\Delta }$ is always equal to

• A. $4$
• B. $2$
• C. $1$
• D. $3$

Answer 1

The correct option is A $4$

$\frac{b^2\sin 2C+c^2\sin 2B}{\Delta }$
$=\frac{2b^2\sin C\cos C+2c^2\sin B\cos B}{\Delta }$
$=\frac{2bc\sin B\cos C+2bc\sin C\cos B}{\Delta }=4$ (using sine rule)