Question

If In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively, and the sides $a,b,c$ are in A.P.

then?

• A. $B>{60}^{\circ }$
• B. $B<{60}^{\circ }$
• C. $B\le {60}^{\circ }$
• D. $B=A-C$

Answer 1

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

Given that, $2b=a+c$
From sine rule: $2\sin B=\sin A+\sin C$
$4\sin \frac{B}{2}\cos \frac{B}{2}=2\sin \left(\frac{A+C}{2}\right)\cos \left(\frac{A-C}{2}\right)$
$2\sin (\frac{\pi }{2}-\frac{A+C}{2})\cos (\frac{\pi }{2}-\frac{A+C}{2})=\sin \left(\frac{A+C}{2}\right)\cos \left(\frac{A-C}{2}\right)$
$\Rightarrow 2\cos \left(\frac{A+C}{2}\right)=\cos \left(\frac{A-C}{2}\right)$
$\Rightarrow \frac{\cos \left(\frac{A+C}{2}\right)}{\cos \left(\frac{A-C}{2}\right)}=\frac{1}{2}$
By componendo-dividendo
$\frac{\cos \left(\frac{A+C}{2}\right)+\cos \left(\frac{A-C}{2}\right)}{\cos \left(\frac{A+C}{2}\right)-\cos \left(\frac{A-C}{2}\right)}=\frac{1+2}{1-2}$
$\Rightarrow \frac{\cos \frac{A}{2}\cos \frac{C}{2}}{\sin \frac{A}{2}\sin \frac{C}{2}}=3$
$\Rightarrow \tan \frac{A}{2}\tan \frac{C}{2}=\frac{1}{3}$ ...(1)
As A.M$\ge$G.M
Therefore, $\frac{\tan \frac{A}{2}+\tan \frac{C}{2}}{2}\ge \sqrt{\tan \frac{A}{2}\tan \frac{C}{2}}=\frac{1}{\sqrt{3}}$ ...(2)

In $\Delta ABC$: $\frac{B}{2}=\frac{\pi }{2}-\left(\frac{A+C}{2}\right)$
$\Rightarrow \cot \frac{B}{2}=\tan \left(\frac{A+C}{2}\right)=\frac{\tan \frac{A}{2}+\tan \frac{C}{2}}{1-\tan \frac{A}{2}\tan \frac{C}{2}}$
$\cot \frac{B}{2}\ge \frac{2/\sqrt{3}}{1-\left(1/3\right)}=\sqrt{3}$ [ from (1) and (2) ]
Therefore, $B\le {60}^{\circ }$
Ans: C