Answer
$V_{app} = 3.24~c$
Work Step by Step
In part (a), we found that $~~D_{app} = vt~sin~\theta$
In part (b), we found that $~~T_{app} = t-\frac{vt~cos~\theta}{c}$
We can find $V_{app}$:
$V_{app} = \frac{D_{app}}{T_{app}}$
$V_{app} = \frac{vt~sin~\theta}{t-\frac{vt~cos~\theta}{c}}$
$V_{app} = \frac{v~sin~\theta}{1-\frac{v~cos~\theta}{c}}$
$V_{app} = \frac{(0.980~c)~sin~30.0^{\circ}}{1-\frac{(0.980~c)~cos~30.0^{\circ}}{c}}$
$V_{app} = \frac{(0.980~c)~sin~30.0^{\circ}}{1-0.980~cos~30.0^{\circ}}$
$V_{app} = 3.24~c$
