Answer
The tension in string $ab$ is $~~15~N$
Work Step by Step
Let $T_{ab}$ be the tension in string $ab$
Let $T_{bc}$ be the tension in string $bc$
The horizontal component of the tension in each string is equal in magnitude:
$T_{bc}~cos~\theta_1 = T_{ab}~cos~\theta_2$
$T_{bc} = \frac{T_{ab}~cos~\theta_2}{cos~\theta_1}$
To find $T_{ab}$, we can consider the vertical forces:
$T_{ab}~sin~\theta_2+Mg - T_{bc}~sin~\theta_1 = 0$
$T_{ab}~sin~\theta_2+Mg - \frac{T_{ab}~cos~\theta_2}{cos~\theta_1}~sin~\theta_1 = 0$
$T_{ab}~sin~\theta_2+Mg - T_{ab}~cos~\theta_2~tan~\theta_1 = 0$
$T_{ab}~cos~\theta_2~tan~\theta_1 - T_{ab}~sin~\theta_2 = Mg$
$T_{ab} = \frac{Mg}{cos~\theta_2~tan~\theta_1 - sin~\theta_2}$
$T_{ab} = \frac{(2.0~kg)(9.8~m/s^2)}{cos~20^{\circ}~tan~60^{\circ} - sin~20^{\circ}}$
$T_{ab} = 15~N$
The tension in string $ab$ is $~~15~N$
