Answer
Exercise 77 with $\theta=4^{\circ}$: $R=644ft$
Exercise 78 with $\theta=4^{\circ}$: $R=1559ft$
Therefore the increasing angle $\theta$ would make $R$ decrease.
Work Step by Step
We are using the formula $R=\frac{V^{2}}{g(f+\tan\theta)}$ where R is the safe radius of the curve, V is the velocity in feet per second, $\theta$ is the superelevation and $f$ and $g$ are constants to calculate the radius of the curve.
In this case taking the values from Exercise 77,
$\theta=4^{\circ}$
$g=32.2$
$f=0.14$
$V=45mph=66ft/s$
Substituting in the formula,
$R=\frac{66^{2}}{32.2(0.14+\tan(4^{\circ}))}$
$R=644ft$ (rounded to the nearest feet)
Taking the values from Exercise 78,
$\theta=4^{\circ}$
$g=32.2$
$f=0.14$
$V=70mph=102.66ft/s$
Substituting in the formula,
$R=\frac{102.66^{2}}{32.2(0.14+\tan(4^{\circ}))}$
$R=1559ft$ (rounded to the nearest feet)
Since the answers for $R$ in Exercises 77 and 78 were smaller, we can conclude that increasing the angle increases the result of the radius curve.
