Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

The value of $(a+b+c)(b+c-a)=kbc$ for which of the following conditions?

• A. $A=90$
• B. $A=45$
• C. $A=30$
• D. $\angle A$ is obtuse

Answer 1

The correct option is A $A=90$

$(a+b+c)(b+c-a)=kbc{b}^2+c^2-a^2=(k-2)bc\frac{b^2+c^2-a^2}{2bc}=\frac{k-2}{2}\left(\frac{k-2}{2}\right)=\cos A$

for $k=2,\cos A=0A=90°$

Hence, $A=90$ is the correct answer.