Calculus for the Practical Man: Chapter 4, Problem 10 (Solve problem without trigonometry possible?) My issue is that I am trying to avoid trigonometry as a means to solve this problem, because he goes into the calculus of trigonometric functions in the subsequent chapter, and, therefore, I think he intends the problem to be solved without trigonometry.
Two automobiles are moving along straight level roads which cross at an angle of sixty degrees, one approaching the crossing at 25 miles an hour and the other leaving it at 30 miles an hour on the same side. How fast are they approaching or separating form each other at the moment when each is ten miles from the crossing.
How do I relate the quantities given that I cannot use trigonometry and the right triangle relation $x^2 + y^2 = h^2$?"
 A: Let's setup the road like a V.
Due to symmetry, both car horizontal velocity roughly cancelled out
Vertically, they are traveling in opposite direction.
Thus, they are moving away from each other: 
(25 + 30)*sin(30°) = 55/2 = 27.5 mph
Update 1:
Removed assumption of horizontal velocity cancellation.
Use law of cosine for more accurate relative speed, 
$v^2 = 25^2 + 30^2 - (2)(25)(30) \cos(60°) = 775$
$v = \sqrt{775} ≈ 27.84 $ mph
Update 2:
I finally figured out what's wrong.
The question does not concern with relative speed, only rate of distance change when both at 10 miles mark.
Just to confirm, say $10^{-6}$ hour passed since the 10 miles mark.
Redo law of cosine, $a = 10-0.000025, b=10+0.000030$
$Δd = \sqrt{a^2 + b^2 - 2 a b \cos(60°)} - 10 ≈ 2.5\times 10^{-6}$ miles   
Rate of distance change ≈ $\frac{Δd}{Δt} = 2.5$ mph
A: It is indeed possible to derive the solution without any trig usage, given the unique geometric configuration in the question.
Observe that the two vehicles and the road-cross form an equilateral triangle. The vehicle moving towards the cross reduces the distance between them while the vehicle moving away from the cross increases the distance.
So, in the special case where both vehicles moved with the same speed, there would be no change of distance due to the equilateral offset argued above. 
In the case where there is a difference in their speeds as given in the question, their distance changes at the rate equal to the net difference of their velocities along the direction of the distance line, i.e. 
$$\frac{30-25}{2}=2.5 \space \text{miles/hour}$$
The factor of half is due to the 60-degree, or equilateral, formation among the roads and their distance, which project only half of their velocities along their distance line. (Note that their velocities along the direction perpendicular to the line of the distance do not contribute to the change of their distance.)
A: Here is the proof, using another form of Law of Cosine
$c^2 =(a-b)^2 + 4ab\sin^2(\frac{C}{2}) $
Let $a=10-25t \text{ , } b=10+30t \text{ , }  C=60° \text{, last term reduced to } ab$
$\begin{align} \frac{Δc}{Δt}
&= \frac{c-10}{t} \times \frac{c+10}{c+10} \cr
&= \frac{c^2-100}{t (c+10)} \cr
&= \frac{(55t)^2 + (10-25t)(10+30t) - 100}{t (c+10)} \cr
&= \frac{50 + 2275t}{c+10} \cr
\end{align}$
When t goes to zero, above reduced to $\frac{50}{20} = 2.5$ mph

Second way is to use calculus:  
$\begin{align}c 
&= \sqrt{(55t)^2 + (10-25t)(10+30t)} \cr
&= \sqrt{100 + 50t + 2275t^2}
\end{align}$ 
$v = {dc \over dt} = {50 + 2275(2t) \over2c}$ 
When t goes to zero, $v = {50\over20} = 2.5$ mph  
