Question

In $ABC$, if $(a+b+c)(b+c-a)=3bc$, then

• A. $A={60}^0$
• B. $B={60}^0$
• C. $C={60}^0$
• D. $A={90}^o$

Answer 1

The correct option is A $A={60}^0$

$(b+c+a)(b+c-a)=(b+c{)}^2-a^2$

$=b^2+c^2+2bc-a^2$ $......\left(1\right)$

Substituting (1) in the given expression, we get

$b^2+c^2-a^2=3bc-2bc$

That gives, $\frac{b^2+c^2-a^2}{2bc}=\frac{1}{2}$ $......\left(2\right)$

We also know, by cosine law that $\cos A=$ $\frac{b^2+c^2-a^2}{2bc}$.

By equation (2), $\cos A=\frac{1}{2}$ or $A={60}^{\circ }$. Option A is correct.