In triangle ABC,if a, b, c are 5, 12 and 13 r | vedclass

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(B) Given: $a=5, b=12, c=13$. Since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$,the triangle is a right-angled triangle with $\angle A = 90^\circ$. Area $\De

Vedclass · Accepted Answer

$120$

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(B) Given: $a=5, b=12, c=13$. Since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$,the triangle is a right-angled triangle with $\angle A = 90^\circ$. Area $\Delta = \frac{1}{2} 	imes a 	imes b = \frac{1}{2} 	imes 5 	imes 12 = 30$. We need to evaluate $b^2 \sin 2C + c^2 \sin 2B$. Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$ and the Sine Rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$: $b^2 (2 \sin C \cos C) + c^2 (2 \sin B \cos B) = 2b^2 \left(\frac{c}{2R}ight) \left(\frac{a^2+b^2-c^2}{2ab}ight) + 2c^2 \left(\frac{b}{2R}ight) \left(\frac{a^2+c^2-b^2}{2ac}ight)$. Simplifying this expression,we get $4\Delta$. Thus,$4 	imes 30 = 120$.

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

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