A body of mass 1 kg is suspended with the help of two strings making angles as shown in figure. Magnitude of tensions $$T_1$$ and $$T_2$$, respectively, are (in N) :

Here as the system is in equilibrium 

We have to equate horizontal components of both $$T_{1\ }and\ T_2$$ as there is no external Force acting in the horizontal direction

Here horizontal component of $$T_{1\ }is\ T_1\cos60^{\circ\ }$$

Here horizontal component of $$T_{2\ }is\ T_2\cos30^{\circ\ }$$

Equate these 2

$$T_2\cos60^{\circ\ }=\ T_1\cos30^{\circ\ }$$

$$T_2\times\ \frac{\sqrt{\ 3}}{2}=\ T_1\times\ \frac{1}{2}$$

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

Now coming to the Vertical direction 

There is gravitational force downwards which is Mg and vertical components of $$T_{1\ }and\ T_2$$ add up and are opposing the force due to gravitation

so $$Mg=\ T_1\sin\theta\ \ +\ T_2\sin\ \ \theta\ $$

$$Mg=\ T_1\times\ \frac{\sqrt{\ 3}}{2}\ +\ T_2\ \times\ \frac{1}{2}$$

Substituting  $$T_1\ =\ \sqrt{\ 3}T_2$$ in that equation

$$Mg=\ \sqrt{\ 3}T_2\times\ \frac{\sqrt{\ 3}}{2}\ +\ T_2\ \times\ \frac{1}{2}$$

$$Mg=\frac{3T_2}{2}+\frac{T_2}{2}$$
$$Mg=2T_2$$

$$T_2=\frac{g}{2}\ \ \left(as\ M=1\right)$$

$$T_2=5N$$

as $$T_1\ =\ \sqrt{\ 3}T_2$$

$$T_1=5\sqrt{\ 3}N$$