Question

In \[\Delta ABC\], if \[\left( {a + b + c} \right)\left( {a - b + c} \right) = 3ac\], then which of the following is correct?
A. \[\angle B = {60^o}\]
B. \[\angle B = {30^o}\]
C. \[\angle C = {60^o}\]
D. \[\angle A + \angle C = {90^o}\]

Answer 1

Hint: Using algebraic identity we will simplify the left side of the given equation. Then using cosine law, we will find which of the given options is correct.

Formula used:
Algebraical identity:
\[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\]
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
Cosine law:
\[{a^2} = {b^2} + {c^2} - 2bc\cos A\]
\[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
\[{c^2} = {a^2} + {b^2} - 2ab\cos C\]

Complete step by step solution:
Given equation is \[\left( {a + b + c} \right)\left( {a - b + c} \right) = 3ac\]
Now applying algebraic identity \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\] on the left side
\[ \Rightarrow {\left( {a + c} \right)^2} - {b^2} = 3ac\]
Now applying algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
\[ \Rightarrow {a^2} + {c^2} + 2ac - {b^2} = 3ac\]
\[ \Rightarrow {a^2} + {c^2} - {b^2} = 3ac - 2ac\]
\[ \Rightarrow {a^2} + {c^2} - {b^2} = ac\]
We know that, \[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
\[ \Rightarrow {a^2} + {c^2} - {b^2} = 2ac\cos B\]
Now substitute \[{a^2} + {c^2} - {b^2} = ac\] in the equation
\[ \Rightarrow ac = 2ac\cos B\]
Cancel out ac from both sides of equation
\[ \Rightarrow 1 = 2\cos B\]
\[ \Rightarrow \cos B = \dfrac{1}{2}\]
We know that \[\cos {60^o} = \dfrac{1}{2}\]
\[ \Rightarrow \cos B = \cos {60^o}\]
\[ \Rightarrow B = {60^o}\]
Hence option A is the correct option.

Additional information:
Some formulas are there related to the triangle, that is sin law
According to the sine law \[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\]

Note: Students often confuse with value of \[\cos {60^o}\] and \[\cos {30^o}\]. The value of \[\cos {60^o}\] is \[\dfrac{1}{2}\] and \[\cos {30^o}\] is \[\dfrac{{\sqrt 3 }}{2}\].