Answer
The length of the third side is increasing at a rate of $~~0.3958~m/min$
Work Step by Step
Let $a, b,$ and $c$ be the three sides of the triangle.
We can find $c$ when $a = 12~m$, $b = 15~m$, and $\theta = \frac{\pi}{3}~rad$:
$c^2 = a^2+b^2-2ab~cos~\theta$
$c = \sqrt{a^2+b^2-2ab~cos~\theta}$
$c = \sqrt{12^2+15^2-2(12)(15)~cos~\frac{\pi}{3}}$
$c = \sqrt{189}$
We can express $\frac{d\theta}{dt}$ in units of $rad/min$:
$\frac{d\theta}{dt} = (2^{\circ}/min)(\frac{\pi~rad}{180^{\circ}}) = \frac{\pi}{90}~rad/min$
We can find $\frac{dc}{dt}$:
$2c~\frac{dc}{dt} = 2a~\frac{da}{dt}+2b~\frac{db}{dt}-2b~cos~\theta~\frac{da}{dt}-2a~cos~\theta~\frac{db}{dt}+2ab~sin~\theta~\frac{d\theta}{dt}$
$\frac{dc}{dt} = \frac{1}{c}~[a~(0)+b~(0)-b~cos~\theta~(0)-a~cos~\theta~(0)+ab~sin~\theta~\frac{d\theta}{dt}]$
$\frac{dc}{dt} = \frac{1}{c}~(ab~sin~\theta~\frac{d\theta}{dt})$
$\frac{dc}{dt} = \frac{(12)(15)~(sin~\frac{\pi}{3})(\frac{\pi}{90})}{\sqrt{189}}$
$\frac{dc}{dt} = 0.3958~m/min$
The length of the third side is increasing at a rate of $~~0.3958~m/min$
