Question

The angles $\alpha ,\beta ,\gamma$ of a triangle are in A.P. If the greatest angle $\gamma$ is double the least angle $\alpha$, then

• A. $\alpha :\beta =2:3$
• B. $\alpha :\beta :\gamma =2:3:5$
• C. $\beta :\gamma =3:4$
• D. $sin\left(3\alpha \right)=cos\left(\frac{\beta }{2}\right)$

Answer 1

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

The correct options are
A $\alpha :\beta =2:3$
C $\beta :\gamma =3:4$
D $sin\left(3\alpha \right)=cos\left(\frac{\beta }{2}\right)$

Let the angles be $a-d,a,a+d$
Since $(a-d)+\left(a\right)+(a+d)={180}^0$
$3a={180}^0$ $⟹a={60}^0$
The angles are $60-d,60,60+d$
Also, $2(60-d)=60+d$
$120-2d=60+d⟹{60}^0=3d,⟹d={20}^0$
Required angles are ${40}^0,{60}^0,{80}^0$
Hence, $\alpha :\beta :\gamma =2:3:4$
Also,
$\sin \left(3\alpha \right)=\sin 120=\frac{\sqrt{3}}{2}=\cos \left(\frac{60}{2}\right)$
Hence, options A, and D are correct