Answer
(a) $\theta = 23^{\circ}$
(b) The angle is toward the windshield.
Work Step by Step
(a) First let's find the acceleration.
$a = \frac{v-v_0}{t} = \frac{0-25 ~m/s}{6.0 ~s} = -4.2 ~m/s^2$
The magnitude of acceleration is $a = 4.2 ~m/s^2$.
We can use the acceleration to find the angle.
$F_T ~sin(\theta) = ma$
$F_T ~cos(\theta) = mg$
Let's divide the first equation by the second equation.
$tan(\theta) = \frac{a}{g}$
$\theta = tan^{-1}(\frac{a}{g}) = tan^{-1}\frac{4.2 ~m/s^2}{9.80 ~m/s^2} = 23^{\circ}$
(b) The angle is toward the windshield. The car is decelerating, so the horizontal component of the tension is pulling the object backward. Therefore the angle must be toward the windshield.
