Question

If the angles $A<B<C$ of a $∆ABC$ are in A. P. then

• A. $c^2=a^2+b^2–a{b}^2$
• B. $c^2=a^2+b^2$
• C. $a^2=b^2+c^2–bc$
• D. $b^2=a^2+c^2–a{c}^2$

Answer 1

The correct option is D

$b^2=a^2+c^2–a{c}^2$

Explanation for Correct Option:

Step 1: The Sum of all angles of the triangle is $180°$.

So, $A+B+C={180}^{°}.....\left(i\right)$

Since $A,B,C$ are in A.P.

So, $2B=A+C.....\left(\mathrm{ii}\right)$

Step 2: Putting the value of equation (ii) in equation (i) We get,

$\therefore 3B={180}^{°}orB={60}^{°}$

Now,

$$\begin{gathered} \Rightarrow b^2=a^2+c^2-2ac\cos B \\ \Rightarrow b^2=a^2+c^2-2ac\cos 60° \\ \Rightarrow b^2=a^2+c^2-ac \end{gathered}$$

Hence option (3) is Correct