Problem 16

Given the parametric equations of the line:
\begin{cases}
x=-2+t\cos(\text{30°}) \\
y=3-t\sin(\text{60°})
\end{cases}

To find the angle of inclination of the line, we need to determine its slope ($m$). From the parametric equations, we obtain the change in $x$ and $y$ with respect to the parameter $t$:

\begin{align*}
\Delta x &= \cos(\text{30°}) \\
\Delta y &= -\sin(\text{60°})
\end{align*}

Therefore, the slope $m$ of the line is the ratio of $\Delta y$ to $\Delta x$:

\begin{align*}
m &= \frac{\Delta y}{\Delta x} \\
m &= \frac{-\sin(\text{60°})}{\cos(\text{30°})} \\
m &= \frac{-\sqrt{3}/2}{\sqrt{3}/2} \\
m &= -1
\end{align*}

The slope of the line is equal to the tangent of its angle of inclination $\theta$ with the positive direction of the x-axis, thus:
$$m = -1 = \tan(\theta)$$

To find the angle $\theta$ that corresponds to $\tan(\theta) = -1$, we can consider angles in the second quadrant where the tangent is negative, and we know that $\tan(135°) = -1$.

Therefore, we conclude that the angle of inclination of the line is $\boxed{\text{135°}}$.