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 $x^2 + y^2 = h^2$?"
Best Answer
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