Question

If the angles of a triangle are in arithmetic progression such that $\sin (2A+B)=\frac{1}{2}$, then

• A. $A={45}^o$
• B. $C={75}^o$
• C. $\sin (B+2C)=\frac{-1}{2}$
• D. $\cos 2B=\frac{-1}{2}$

Answer 1

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

The correct options are
A $A={45}^o$
B $C={75}^o$
C $\sin (B+2C)=\frac{-1}{2}$
D $\cos 2B=\frac{-1}{2}$

A, B, C are in A.P $\Rightarrow 2B=A+C$
also for triangle, $A+B+C={180}^{\circ }$
$\Rightarrow B={60}^{\circ }$
also given, $\sin (2A+B)=\frac{1}{2}\Rightarrow 2A+B={150}^{\circ }$
solving above equations,
$A={45}^{\circ },B={60}^{\circ },C={75}^{\circ }$
$\Rightarrow \sin (B+2C)=\sin {210}^{\circ }=-\frac{1}{2},\cos 2B=\cos {120}^{\circ }=-\frac{1}{2}$
Hence options A, B, C, D are correct.