A body of mass m is suspended by two strings making angles $$\theta_1$$ and $$\theta_2$$ with the horizontal ceiling with tensions $$T_1$$ and $$T_2$$ simultaneously. $$T_1$$ and $$T_2$$ are related by $$T_1 = \sqrt{3}T_2$$. The angles $$\theta_1$$ and $$\theta_2$$ are

Given:

$$T_1=\sqrt{\ 3}T_2$$

Horizontal equilibrium

$$T_1\cos\theta_1=T_2\cos\theta_2$$

Substitute:

$$\sqrt{3}T_2\cos\theta_1=T_2\cos\theta_2$$​

Cancel $$T_2$$​:

$$\sqrt{3}\cos\theta_1=\cos\theta_2$$

From the options

We know:

• $$\cos60^{\circ}=\frac{1}{2}$$
• $$\cos30^{\circ}=\frac{\sqrt{3}}{2}$$

Try:

$$\theta_1=60^{\circ\ }$$

Then:

$$\sqrt{3}\cos60^{\circ}=\sqrt{3}\cdot\frac{1}{2}=\frac{\sqrt{3}}{2}$$​​

So:

$$\cos\ \theta\ _2=\frac{\sqrt{\ 3}}{2}$$

$$\theta\ _2=30^{\circ\ }$$

$$T_1\sin\theta_1+T_2\sin\theta_2=mg$$

Substitute:

$$\sqrt{3}T_2\cdot\frac{\sqrt{3}}{2}+T_2\cdot\frac{1}{2}=mg$$

$$\frac{3}{2}T_2+\frac{1}{2}T_2=mg\Rightarrow2T_2=mg$$

$$T_2 = \dfrac{mg}{2}$$