Question

If the angle of the triangle $ABC$ are in A.P, the $a^2+c^2-ac=\_\_\_\_\_\_\_$

Answer 1

Step 1: Use angle sum property of triangle

Let the three Angles of triangle $ABC$ in A.P as

$\angle A=a-d$

$\angle B=a$

$\angle C=a+d$

Since, sum of angles of a triangle is ${180}^0$

$$\begin{gathered} \Rightarrow \angle A+\angle B+\angle C=180° \\ \Rightarrow a-d+a+a+d=180° \\ \Rightarrow 3a=180° \\ \Rightarrow a=60° \end{gathered}$$

Therefore $\angle B=60°$

Step 2: Use cosine formula and find the required result

As per cosine formula

$$\begin{gathered} \cos \left(B\right)=\frac{a^2+c^2-b^2}{2ac} \\ \Rightarrow \cos \left(60°\right)=\frac{a^2+c^2-b^2}{2ac} \\ \Rightarrow \frac{1}{2}=\frac{a^2+c^2-b^2}{2ac} \\ \Rightarrow 1=\frac{a^2+c^2-b^2}{ac} \\ \Rightarrow ac=a^2+c^2-b^2 \\ \Rightarrow a^2+c^2-ac=b^2 \end{gathered}$$

Hence, $a^2+c^2-ac=b^2$