If the angles of a triangle ABC be in A.P., then - Vidyadip

MCQ

If the angles of a triangle ABC be in A.P., then

A
$c^2=a^2+b^2-a b$
✓
$b^2=a^2+c^2-a c$
C
$a^2=b^2+c^2-a c$
D
$b^2=a^2+c^2$

✓

Answer

Correct option: B.

$b^2=a^2+c^2-a c$

(B) $A , B , C$ are in A . P. then angle $B =60^{\circ}$, $\ldots\left[\begin{array}{l}\because A+B+C=180^{\circ} \\ \Rightarrow A+C=2 B \Rightarrow B=60^{\circ}\end{array}\right]$
$\therefore \quad \cos B=\frac{a^2+c^2-b^2}{2 a c}$,
$\Rightarrow \frac{1}{2}=\frac{ a ^2+ c ^2- b ^2}{2 ac } \Rightarrow a ^2+ c ^2- b ^2= ac$
$\Rightarrow b ^2= a ^2+ c ^2- ac$

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