Question

In an acute angles triangle $ABC$ if $\cos A,1-\cos B,\cos C$ are in $A.P.$ and $\sin A+\sin C=1$, then $\angle B$ is equal to

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. None of these

Answer 1

The correct option is C $\frac{\pi }{6}$

Given $\sin A+\sin C=1$

$\cos A+\cos C=2(1-\cos B)$

we know that $A+B+C=\pi$

$\Rightarrow 2\cos \left(\frac{A+C}{2}\right)\cos \left(\frac{A-C}{2}\right)=4{\sin }^2\frac{B}{2}$

$\Rightarrow 2\cos \left(\frac{\pi -B}{2}\right)\cos \left(\frac{A-C}{2}\right)-4{\sin }^2\frac{B}{2}=0$

$\Rightarrow \sin \frac{B}{2}[\cos \left(\frac{A-C}{2}\right)-2\sin \left(\frac{\pi -(A+C)}{2}\right)]=0$ since $2\cos c\cos d=\cos (c+d)-\cos (c-d)$

$\Rightarrow \sin \frac{B}{2}[\cos \left(\frac{A-C}{2}\right)-2\cos \left(\frac{A+C}{2}\right)]=0$

as $\sin \frac{B}{2}\ne 0$ so $\cos \left(\frac{A-C}{2}\right)=2\cos \left(\frac{A+C}{2}\right)$

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

multiplying both sides by $2\cos \frac{B}{2}$, we get

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

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

$\Rightarrow \sin A+\sin C=2\sin B$ since $2\sin c\cos d=\sin (c+d)+\sin (c-d)$

$\Rightarrow \sin B=\frac{1}{2}\Rightarrow \angle B=\frac{\pi }{6}$