Question

If $A,B,C$ are the three angles in a triangle such that $2\sin B\sin (A+B)-\cos A=1$ and $2\sin C\sin (B+C)-\cos B=0$, then

• A. $A={120}^{\circ }$
• B. $B={120}^{\circ }$
• C. $C={30}^{\circ }$
• D. $B=C$

Answer 1

Answer: (A), (C), (D)

$B=C$

$2\sin B\sin C-\cos A=1$

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

$\cos (B-C)=1\Rightarrow B=C$

$2\sin B\sin \left(2B\right)=\cos B$

$\left(2\sin B\right)\left(2\sin B\cos B\right)=\cos B$

$\cos B\ne 0,{\sin }^{2}B=\frac{1}{4}\Rightarrow \sin B=\frac{1}{2}$

$B={30}^{\circ }$

$C={30}^{\circ }$

$A={120}^{\circ }$