Question

In $△ABC$ if $a=13,b=14,c=15$, find the value of $\sin A$.

Answer 1

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To find : sin A

By applying cosine Rule

$cosA=\frac{b^2+c^2-a^2}{2bc}$

$cosA=\frac{196+225-169}{2\left(14\right)\left(15\right)}$

$cosA=\frac{252}{420}=\frac{3}{5}$

$si{n}^{2}A+co{s}^{2}A=1$

$co{s}^{2}A=1-si{n}^{2}A$

$\Rightarrow cosA=\sqrt{1-si{n}^2\theta }$

$\sqrt{1-si{n}^{2}A}=\frac{3}{5}$

Squaring both sides

$1-si{n}^{2}A=\frac{9}{25}$

$25-25si{n}^{2}A=9$

$si{n}^{2}A=\frac{16}{25}$

$sinA=\frac{4}{5}$