(A) Given sides are $a=13, b=14, c=15$. Semi-perimeter $s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = \frac{42}{2} = 21$. Using the formula $\sin(\frac{A

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(A) Given sides are $a=13, b=14, c=15$. Semi-perimeter $s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = \frac{42}{2} = 21$. Using the formula $\sin(\frac{A}{2}) = \sqrt{\frac{(s-b)(s-c)}{bc}}$: $s-b = 21-14 = 7$ $s-c = 21-15 = 6$ $\sin(\frac{A}{2}) = \sqrt{\frac{7 \times 6}{14 \times 15}} = \sqrt{\frac{42}{210}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}$.

Class 11 Mathematics Trigonometrical Equations Relation between sides and angles, Solutions of triangles

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In a $\triangle ABC$,if $a=13, b=14$ and $c=15$,then $\sin(\frac{A}{2})$ is:

AP EAMCET 2022 All AP EAMCET

A
$\frac{1}{\sqrt{5}}$
B
$\frac{1}{\sqrt{7}}$
C
$\frac{3}{4}$
D
$\frac{4}{5}$

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In a $\triangle ABC$,if $a=13, b=14$ and $c=15$,then $\sin(\frac{A}{2})$ is:

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