Question

If in a $△ABC,\cos A=\frac{\sin B}{2\sin C}$, then it is

• A. An isosceles triangles
• B. An equilateral triangle
• C. A right angled triangle
• D. None of these

Answer 1

The correct option is A An isosceles triangles

In $\Delta$ABC,

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k$(say)

So,

$\sin A=ka$

$\sin B=kb$

$\sin C=kc$

Where a, b, c are sides of the triangle and cosine formula is given by

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

Now the expression is

$\cos A=\frac{\sin B}{2\sin C}$

putting all the value we get

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

or $b^2+c^2-a^2=b^2$

$c=a$

$\therefore$ The triangle is an isosceles.