In triangle ABC, if a = 13, b = 14, c = 15, t | vedclass

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(B) $1$. Calculate the semi-perimeter $s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = 21$.$2$. Calculate the area of the triangle using Heron's formula: $

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$\frac{7}{5}$

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(B) $1$. Calculate the semi-perimeter $s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = 21$.$2$. Calculate the area of the triangle using Heron's formula: $	ext{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 	imes 8 	imes 7 	imes 6} = \sqrt{7056} = 84$.$3$. Use the area formula $	ext{Area} = \frac{1}{2}bc \sin A$ to find $\sin A$: $84 = \frac{1}{2} 	imes 14 	imes 15 	imes \sin A \implies 84 = 105 \sin A \implies \sin A = \frac{84}{105} = \frac{4}{5}$.$4$. Since $\sin^2 A + \cos^2 A = 1$, $\cos A = \sqrt{1 - (\frac{4}{5})^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.$5$. Therefore, $\sin A + \cos A = \frac{4}{5} + \frac{3}{5} = \frac{7}{5}$.

In $\triangle ABC$, if $a = 13, b = 14, c = 15$, then the value of $\sin A + \cos A$ is...

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