Given,

Angle of inclination, $$\theta$$ = 30$$^\circ$$

Acceleration, a = 10 ms^$$-$$2

Acceleration due to gravity, g = 10 ms^$$-$$2

According to the question the car and bob is as shown below,

Here, F' is the pseudo force acting on the bob when we considered it from car's frame and T is the tension on the string.

In equilibrium, $$\Sigma$$F_x = 0 and $$\Sigma$$F_y = 0

$$\Rightarrow$$ F' cos30$$^\circ$$ = T sin$$\alpha$$

$${{ma\cos 30^\circ } \over {\sin \alpha }} = T$$ ...... (i)

where, m is the mass of the bob.

$$F'\sin 30^\circ + mg = T\cos \alpha $$

$$ \Rightarrow ma\sin 30^\circ + mg = {{ma\cos 30^\circ } \over {\sin \alpha }}(\cos \alpha )$$ [$$\because$$ using Eq. (i)]

$$ \Rightarrow a\sin 30^\circ + g = {{a\cos 30^\circ } \over {\sin \alpha }}(\cos \alpha )$$

$$10 \times {1 \over 2} + 10 = {{10 \times \sqrt 3 } \over 2}\cot \alpha $$

$$ \Rightarrow {1 \over 2} + 1 = {{\sqrt 3 } \over 2}\cot \alpha \Rightarrow {3 \over 2} = {{\sqrt 3 } \over 2}\cot \alpha $$

or, $$\cot \alpha = \sqrt 3 $$

$$ \Rightarrow \alpha = 30^\circ $$