Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 35

At what rate is the length of the third side increasing when the angle between the sides of fixed length is $60^\circ$

$
\begin{equation}
\begin{aligned}
\text{Given: } L_1 &= 12 m\\
L_2 &= 15m\\
\\
\frac{d \theta}{dt} &= 2\frac{^\circ}{\text{min}} \left( \frac{\pi}{180} \right) = \frac{2\pi}{180} /\text{min}
\end{aligned}
\end{equation}
$

Required: $\displaystyle \frac{dx}{dt}$ when $\theta = 60^{\circ}$




We will use cosine law for triangles to solve $x$ at $\theta = 60^\circ$

$
\begin{equation}
\begin{aligned}
x^2 &= L_1^2 + L_2^2 - 2(L_1)(L_2) \cos \theta && \Longleftarrow \text{ Equation 1}\\
\\
x &= \sqrt{12^2 + 15^2 - 2(12)(15) \cos 60^{\circ}}\\
\\
x &= 3\sqrt{21}
\end{aligned}
\end{equation}
$

We will derive the Equation 1 to solve for $\displaystyle \frac{dx}{dt}$, given that $L_1$ and $L_2$ are constants

$
\begin{equation}
\begin{aligned}
\cancel{2}\frac{dx}{dt} &= \cancel{2}(L_1)(L_2) \sin \theta \left( \frac{d \theta}{dt} \right)\\
\\
\frac{dx}{dt} &= \frac{(L_1)(L_2)\sin \theta \left( \frac{d \theta}{dt}\right)}{x}\\
\\
\frac{dx}{dt} &= \frac{(12)(15)\sin 60\left( \frac{2\pi}{180}\right)}{3\sqrt{21}}
\end{aligned}
\end{equation}\\
\quad\boxed{\displaystyle \frac{dx}{dt} = 0.396 \frac{m}{\text{min}}}
$