If the angles $A,\ B,\ C$ of a $\vartriangle ABC$ are in A.P., then prove that $b^2=a^2+c^2-ac$.
Given: Angles $A,\ B,\ C$ of a $\vartriangle ABC$ are in A.P.
To do: To prove that $b^2=a^2+c^2-ac$.
Solution:
We have $A+B+C=180^o$
Also $A,\ B,\ C$ are in A.P.
$\Rightarrow 2B=A+C$
$\therefore 3B=180^o$
$\Rightarrow B=60^o$
Now $b^2=a^2+c^2−2ac.cosB$
$=a^2+c^2−2ac.cos60^o$
$\Rightarrow b^2=a^2+c^2−ac$
Hence proved.
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