Question

If in $△ABC,$ sides $a,b,c$ are in A.P. then :

• A. $B>{60}^o$
• B. $B<{60}^o$
• C. $B\le {60}^o$
• D. None of the these

Answer 1

The correct option is C $B\le {60}^o$

It is given that, in $△ABC,$ $a,b,c$ are in A.P.

$\therefore$ $2b=a+c$

$\Rightarrow$ $2\sin B=\sin A+\sin C$

$\Rightarrow$ $2\times 2\sin \frac{B}{2}\cos \frac{B}{2}=2\sin \left(\frac{A+C}{2}\right)\cos \left(\frac{A-C}{2}\right)$

Here, $\sin \left(\frac{A+C}{2}\right)=\sin (\frac{\pi }{2}-\frac{B}{2})$

$\Rightarrow$ $2\sin \frac{B}{2}\cos \frac{B}{2}=\sin (\frac{\pi }{2}-\frac{B}{2})\cos \left(\frac{A-C}{2}\right)$

$\Rightarrow$ $2\sin \frac{B}{2}\cos \frac{B}{2}=\cos \left(\frac{B}{2}\right)\cos \left(\frac{A-C}{2}\right)$

$\Rightarrow$ $2\sin \frac{B}{2}=\cos \left(\frac{A-C}{2}\right)$

$\Rightarrow$ $\sin \frac{B}{2}=\frac{1}{2}\cos \left(\frac{A-C}{2}\right)$

Maximum value of $\cos \left(\frac{A-C}{2}\right)=1$

$\Rightarrow$ $\sin \frac{B}{2}\le \frac{1}{2}$

$\Rightarrow$ $\sin \left(\frac{B}{2}\right)\le \sin \left(\frac{\pi }{6}\right)$

$\Rightarrow$ $\frac{B}{2}\le \frac{\pi }{6}$

$\Rightarrow$ $B\le \frac{\pi }{3}$

$\Rightarrow$ $B\le {60}^o$