Question

In $△ABC,a=3,b=4$ and $c=5$, then value of $\sin A+\sin 2B+\sin 3C$ is-

• A. $\frac{39}{25}$
• B. $\frac{12}{25}$
• C. $\frac{14}{25}$
• D. None of these

Answer 1

Answer: (C)

$\frac{14}{25}$

$△=\sqrt{s(s-a)(s-b)(s-c)}=6$
$\cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{-16+9+25}{30}=\frac{18}{30}$

$\sin A=\frac{2△}{bc}=\frac{3}{5}$

$\sin B=\frac{2△}{ac}=\frac{12}{15}=\frac{4}{5}$

$\sin C=\frac{2△}{ab}=\frac{12}{12}=1$
Now $\sin A+\sin 2B+\sin 3C$
$=\sin A+2\sin B\cos B+3\sin C-4{\sin }^{3}C=\frac{14}{25}$