Question

If in a triangle $ABC,2\cos A=\sin BcosecC$, then

• A. $a=h$
• B. $b=c$
• C. $c=a$
• D. $2a=bc$

Answer 1

The correct option is C $c=a$

As we know that

$\frac{\sin B}{\sin C}=\frac{b}{c}$

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

Given $2\cos A=\sin B\text{cosec}C$

$⟹2\cos A=\frac{\sin B}{\sin C}$

$⟹2\times \frac{b^2+c^2-a^2}{2bc}=\frac{b}{c}$

$⟹b^2+c^2-a^2=b^2$

$⟹c^2=a^2$

$⟹c=a$