Q:

The rate of change of the distance from the plane to the radar station

Problem statement:
A plane flying with constant speed of 4 km/min passes over a ground radar station at an altitude of 6 km and climbs at an angle of 35 degrees. At what rate, in km/min, is the distance from the plane to the radar station increasing 6 minutes later? 
Hint: The law of cosines for a triangle is $c^2=a^2+b^2− 2ab \cos(\theta)$ where $\theta$ is the angle between the sides of length $a$ and $b$.

Okay, so following the typical related rates algorithm:
Identify the problem at hand by drawing a picture. I did that. It appears to me that after the plane passes over the ground station, we get an angle between the measures a and b of 90 + 35 degrees. 
Awesome.
Now, let's build a table of values:
$$
a = 6\,\mathrm{km}\\
a' = 0 \quad \text{(duh)}\\
b = b'\cdot t = 4\cdot 6 = 24\\
b' = 4\,\mathrm{km}/\mathrm{min}\\
t' = 6min\\
c = \sqrt{a^2 + b^2} = \sqrt{36+24} = 2\sqrt{15} \approx 2.78316\\
c' = ?
$$
Okay, let's differentiate the law of cosines, which is given to us in form of a hint:
$2cc' = 2aa' + 2bb' - \frac{du}{dt}$
$
u = 2ab\cos(\theta)$
$\frac{du}{dt} = \frac{ds}{dt} \cdot \cos(\theta) - \sin(\theta)\cdot(2ab)$
$\frac{ds}{dt} = 0\cdot ab + (a'b + b'a)$
And so now we rewind:
$\frac{du}{dt} = \left[ \left(0 \cdot ab + \left(a'b + b'a \right) \right) \right] \cdot \cos(\theta) - \sin(\theta) \cdot 2ab $
And so: 
$2cc' = 2aa' + 2bb' - \left[ \left(0 \cdot ab + \left(a'b + b'a \right) \right) \right] \cdot \cos(\theta) - \sin(\theta) \cdot 2ab $

Boom, boom. Let's substutute what we know
$2\left(2 \sqrt{15} \right)(c') = 2(6)(0) + 2(24)(4) - \left[ \left(0\cdot (6)(24) + \left((0)(24) +  (4)(6) \right) \right) \right] \cdot \cos(125^\circ) - \sin(125^\circ) \cdot 2(4)(24)$

Now what? Well, evaluate. Not sure if that's going to yield the right answer. But I don't think that it will. 
Please help! I don't know how to solve this kind of problem! It seems my solution/process is incorrect!

A:

We use your notation. Let $c=c(t)$ be the distance from the plane to the radar station at time $t$. Let $b$ be the distance from the plane to the point where the climb began. Then
$$c^2=36+b^2-12b\cos \theta=36+b^2+12b\sin\phi,\tag{1}$$
where $\theta=125^\circ$ and $\phi=35^\circ$. Differentiating, we get
$$2c\frac{dc}{dt}=2b\frac{db}{dt}+12\sin\phi\frac{db}{dt}.\tag{2}$$
"Freeze" at time $6$ minutes after the climb began. We know everything on the right-hand side of (2), since at that instant $b=(6)(4)$.
We need to calculate $c$ at time $t=6$. Equation (1) does that, with $b=24$. 

