Question

lf in a triangle $ABC$, 3 $\sin A=6\sin B=2\sqrt{3}\sin C$, then the angle $A$ is

• A. ${30}^0$
• B. ${45}^0$
• C. ${60}^0$
• D. ${90}^0$

Answer 1

The correct option is D ${90}^0$

$a=k\sin A$
$L=k\sin B$
$c=k\sin C$
$\sin A=\frac{K^{{}^{\prime }}}{3}$
$\sin B=\frac{K^{{}^{\prime }}}{6}$
$\sin C=\frac{K^{{}^{\prime }}}{2\sqrt{3}}a=\frac{K^{{}^{\prime \prime }}}{3},L=\frac{K^{{}^{\prime \prime }}}{6},c=\frac{K^{{}^{\prime \prime }}}{2\sqrt{3}}$
$\cos A=\frac{L^2+c^2-a^2}{2Lc}$
$=\frac{\frac{1}{6^2}+\frac{1}{{\left(2\sqrt{3}\right)}^2}-\frac{1}{3^2}}{2\times \frac{1}{6}\times \frac{1}{2\sqrt{3}}}$
$=0$
$\cos A=0$
$A={90}^{\circ }$