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(C) Step $1$: Calculate the semi-perimeter $s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = 21$.Step $2$: Use the formula $\sin(\frac{A}{2}) = \sqrt{\frac{

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$\frac{5 + 4\sqrt{5}}{13}$

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(C) Step $1$: Calculate the semi-perimeter $s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = 21$.Step $2$: Use the formula $\sin(\frac{A}{2}) = \sqrt{\frac{(s-b)(s-c)}{bc}} = \sqrt{\frac{(21-14)(21-15)}{14 	imes 15}} = \sqrt{\frac{7 	imes 6}{14 	imes 15}} = \sqrt{\frac{42}{210}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}$.Step $3$: Calculate $\cos(\frac{A}{2}) = \sqrt{\frac{s(s-a)}{bc}} = \sqrt{\frac{21(21-13)}{14 	imes 15}} = \sqrt{\frac{21 	imes 8}{210}} = \sqrt{\frac{168}{210}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$.Step $4$: Calculate $\sin A = 2 \sin(\frac{A}{2}) \cos(\frac{A}{2}) = 2 	imes \frac{1}{\sqrt{5}} 	imes \frac{2}{\sqrt{5}} = \frac{4}{5}$.Step $5$: Sum = $\sin(\frac{A}{2}) + \sin A = \frac{1}{\sqrt{5}} + \frac{4}{5} = \frac{\sqrt{5} + 4}{5} = \frac{5 + 4\sqrt{5}}{13}$ (Wait, correcting sum: $\frac{\sqrt{5}}{5} + \frac{4}{5} = \frac{4 + \sqrt{5}}{5}$). Since the options provided were inconsistent, the correct value is $\frac{4 + \sqrt{5}}{5}$.

In $\triangle ABC$, with usual notation, if $a = 13, b = 14, c = 15$, then the sum of the values of $\sin(\frac{A}{2})$ and $\sin A$ is....

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