Question

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

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

As $A+B+C=180$
And $A,B,C$ is in A.P
$A+C=2B\Rightarrow 180-B=2B\Rightarrow B=60$
Now
$\sin (2A+B)=\frac{1}{2}\Rightarrow \sin (2A+60)=\frac{1}{2}$
$2A+60=150\Rightarrow 2A=90\Rightarrow A=45$

$A+B+C=180\Rightarrow C=75$

$\sin (B+2C)=\sin (60+150)=\sin 210=\sin (180+30)=-\sin 30=-\frac{1}{2}$

$\cos 2B=\cos \left(120\right)=\cos (90+30)=-\sin 30=-\frac{1}{2}$