In $ riangle A B C$, if $a, b, c$ are 5,12 and 13 respectively, then $b^2 \sin 2 C+c^2 \sin 2 B=$

Question

In $	riangle A B C$, if $a, b, c$ are 5,12 and 13 respectively, then $b^2 \sin 2 C+c^2 \sin 2 B=$

Aicharya · Accepted Answer

Given: $a=5, b=12, c=13$ Thus, $\mathrm{S}=\frac{a+b+c}{2}=15$ $\Delta=\sqrt{s(\mathrm{~s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}=\sqrt{15 	imes 10 	imes 3 	imes 2}=30$ Now, $b^2 \sin 2 C+c^2 \sin 2 B$ $\begin{aligned} & =b^2 \cdot 2 \sin C \cos C+c^2 \cdot 2 \sin B \cos B \ & =2 b^2 \cdot \frac{2 \Delta}{a b} \cdot \frac{a^2+b^2-c^2}{2 a b}+2 c^2 \cdot \frac{2 \Delta}{a c} \cdot \frac{a^2+c^2-b^2}{2 a c} \ & =\frac{2 \Delta}{a^2}\left(\mathrm{a}^2+b^2-c^2ight)+\frac{2 \Delta}{a^2}\left(\mathrm{a}^2+c^2-b^2ight) \ & =\frac{2 \Delta}{a^2}\left(2 a^2ight)=4 \Delta=4 	imes 30=120\end{aligned}$

In $\triangle A B C$, if $a, b, c$ are 5,12 and 13 respectively, then $b^2 \sin 2 C+c^2 \sin 2 B=$

Solution