In a triangle ABC with usual notations,if a=1 | vedclass

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(A) First,calculate the semi-perimeter $s$ of the triangle: $s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = \frac{42}{2} = 21$. Using Heron's formula,the

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$\frac{4}{5}$

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(A) First,calculate the semi-perimeter $s$ of the triangle: $s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = \frac{42}{2} = 21$. Using Heron's formula,the area $\Delta$ is: $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 	imes 8 	imes 7 	imes 6} = \sqrt{7056} = 84$. We know the area formula $\Delta = \frac{1}{2}bc \sin A$. Substituting the values: $84 = \frac{1}{2} 	imes 14 	imes 15 	imes \sin A$. $84 = 105 \sin A$. $\sin A = \frac{84}{105} = \frac{4}{5}$.

In a triangle $ABC$ with usual notations,if $a=13, b=14, c=15$,then $\sin A=$

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