Question

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

• A. $\angle B=60°$
• B. $\angle B=30°$
• C. $\angle C=60°$
• D. $\angle A+\angle C=90°$

Answer 1

The correct option is A

$\angle B=60°$

Explanation for the correct Option:

Determining the angle In $∆ABC$

$$\begin{gathered} \Rightarrow (a+b+c)(a-b+c)=3ac\left[\mathrm{Given}\right] \\ \Rightarrow \left(\right(a+c)+b)\left(\right(a+c)–b)=3ac \\ \Rightarrow {(a+c)}^2–b^2=3ac \\ \Rightarrow a^2+c^2+2ac–b^2–3ac=0 \\ \Rightarrow a^2+c^2–b^2=ac\dots \left(i\right) \end{gathered}$$

We know the cosine rule in $∆ABC,$ having sides $a,b,c$

$\mathbf{cos}\mathbf{}\mathbf{B}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{(}{\mathbf{a}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{c}}^{\mathbf{2}}\mathbf{}\mathbf{–}\mathbf{}{\mathbf{b}}^{\mathbf{2}}\mathbf{)}}{\mathbf{2}\mathbf{ac}}$

Putting $\left(i\right)$ in $\cos B$ we get

$$\begin{gathered} \Rightarrow \cos B=\frac{ac}{2ac} \\ \Rightarrow \cos B=\frac{1}{2} \\ \Rightarrow \cos B=\cos 60° \\ \Rightarrow \angle B=60° \end{gathered}$$

Hence, option A is the correct answer.