Question

If a=13,b=12,c=5 in $\Delta ABC$ then $sinA$= ?

Answer 1

We have,

$a=13$

$b=12$

$c=5$

Then, we know that

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

$\cos A=\frac{{12}^2+5^2-{13}^2}{2\times 12\times 5}$

$\cos A=\frac{144+25-169}{120}$

$\cos A=\frac{169-169}{120}$

$\cos A=\frac{0}{120}$

$\cos A=0$

$\cos A=\cos \frac{\pi }{2}$

$A=\frac{\pi }{2}$

\end{align}$

Then,

$\begin{align}

$\sin A=\sin \frac{\pi }{2}$

$\sin A=1$

Hence, this is the answer.