Component of tension force along the plane \(= T \cos \alpha\)

Component of weight along the plane \(= mg \sin \theta\)

Acceleration of the car = Acceleration of the bob in the car as the bob is in equilibrium in frame of the car

So, net force along the plane

\(m g \sin \theta+T \cos \alpha=m a\)

\(mg \sin \theta+ T \cos \alpha= m (g \sin \theta)\)

\(T \cos \alpha=0\)

But \(T\) is not equal to zero so,

\(\cos \alpha=0\)

Then,

\(\alpha=90^{\circ}\)