The free-body diagram (for the hand straps of mass m) is the view that a passenger
might see if she was looking forward and the streetcar was curving toward the right (so $$\vec a$$ points rightward in the figure). We note that |$$\vec a | =v^2/R$$ where $$v = 16 km/h = 4.4 m/s$$
Applying Newton’s law to the axes of the problem (+x is rightward and +y is upward) we
obtain
$$T\sin\theta=m\dfrac{v^2}{R}$$
$$Tcos\theta=mg$$
We solve these equations for the angle:
$$\theta=tan^{-1}\Bigg(\dfrac{v^2}{Rg}\Bigg)$$
which yields $$\theta = 12^\circ$$.