Question

If $2\sin C\cos A=\sin B$, then $\Delta ABC$ is

• A. Isosceles triangle
• B. equilateral triangle
• C. right angle triangle
• D. none of the options

Answer 1

Answer: (A)

$\ln \Delta ABC,2\sin C\cos A=\sin B$ [given]
$\Rightarrow \sin (C+A)+\sin (C-A)=\sin B$
$[\because 2\sin A\cos B-\sin (A+B)+\sin (A-B)]$
$\Rightarrow \sin \left(180^{\circ }-B\right)+\sin (C-A)=\sin B$
$\left[\because \angle A+\angle B+\angle C=180^{\circ }\right]$
$\Rightarrow \sin B+\sin (C-A)=\sin B$
$\Rightarrow \sin (C-A)=0=\sin 0^{\circ }$
$\Rightarrow C-A=0$
$\Rightarrow C=A$
Hence, $△ABC$ is an isosceles triangle.