See Figure 35.

Figure 35

The area $A$ of a triangle with sides $a$ and $b$ and included angle $\theta$ is given by $$A=\dfrac{1}{2}ab\;\sin\;\theta$$

Solution  (a) $\dfrac{\partial A}{\partial a}=\dfrac{1}{2}b\;\sin \theta$ (b) $\dfrac{\partial A}{\partial b}=\dfrac{1}{2} a\;\sin \theta$ (c) $\dfrac{\partial A}{\partial \theta }=\dfrac{1}{2}ab\;\cos \theta$ (d) When $a=20$, $b=30$, and $\theta =\dfrac{\pi }{6}$, then $$\begin{eqnarray*} \dfrac{\partial A}{\partial a} &=&\dfrac{1}{2}(30)\;\left( \sin \frac{\pi }{6}\right) =\dfrac{15}{2} \\[4pt] \dfrac{\partial A}{\partial b} &=&\dfrac{1}{2}(20)\;\left( \sin \frac{\pi }{6}\right) =5 \\[4pt] \dfrac{\partial A}{\partial \theta } &=&\dfrac{1}{2}\left( 20\right) (30)\;\cos \dfrac{\pi }{6}=150\sqrt{3} \end{eqnarray*}$$

(e) Each partial derivative equals the rate of change of area with respect to $a,$ $b,$ or $\theta$. The relative size of each rate of change tells us that the area will increase most rapidly when $a$ and $b$ are fixed and $\theta$ varies. The area increases the least when $a$ and $\theta$ are fixed and $b$ varies.