Hello, mathshelpneeded!
This is not an easy one . . .
Point $\displaystyle M$ lies 500 m north of point $\displaystyle N$.
Cyclist $\displaystyle A$ starts from $\displaystyle M$, travelling south at 6 m/s.
Cyclist $\displaystyle B$ starts from $\displaystyle N$, moving east at 8 m/s.
Let $\displaystyle D$ be the distance between the cyclists.
Let $\displaystyle \angle ABN = x$
a) Show that: .$\displaystyle D \:= \:\frac{4000}{8\sin x + 6\cos x}$ . . Why didn't they reduce this?
b) Calculate the shortest distance between the cyclists. Code:
M*

6t 

A*
 \
 \
5006t  \D
 \
 x \
N*      *B
8t
Cyclist $\displaystyle A$ starts at $\displaystyle M$ and goes south at 6 m/s.
In $\displaystyle t$ seconds, he has gone $\displaystyle 6t$m to point $\displaystyle A$.
$\displaystyle AN \:= \:5006t$
Cyclist $\displaystyle B$ starts at $\displaystyle N$ and goes east at 8 m/s.
In $\displaystyle t$ seconds, he has gone $\displaystyle 8t$m to point $\displaystyle B$.
We want the problem in terms of $\displaystyle x$, not $\displaystyle t$.
In right triangle $\displaystyle ANB\!:\;\;\tan x \:=\:\frac{5006t}{8t}$
We have: .$\displaystyle 8t\tan x \:=\:500  6t\quad\Rightarrow\quad 8t\tan x + 6t \:=\:500$
Factor: .$\displaystyle 2t(\tan x + 3) \:=\:500\quad\Rightarrow\quad t \;=\:\frac{250}{4\tan x + 3} $ [1]
We also have: .$\displaystyle \cos x \:=\:\frac{8t}{D}\quad\Rightarrow\quad D \:=\:\frac{8t}{\cos x} $ [2]
Substitute [1] into [2]: .$\displaystyle D \:=\:\frac{8\left(\frac{250}{4\tan x + 3}\right)}{\cos x} \:=\:\boxed{\frac{2000}{4\sin x + 3\cos x}}$ (a)
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We have: .$\displaystyle D \:=\:2000(4\sin x + 3\cos x)^{1}$
Then: .$\displaystyle D' \;= \;2000(1)(4\sin x + 3\cos x)^{2}(4\cos x  3\sin x) \;=\;0$
We have: .$\displaystyle \frac{2000(3\sin x  4\cos x)}{(4\sin x + 3\cos x)^2} \:=\:0\quad\Rightarrow\quad3\sin x4\cos x \:=\:0$
Then: .$\displaystyle 3\sin x \:=\:4\cos x\quad\Rightarrow\quad\frac{\sin x}{\cos x} \:=\:\frac{4}{3}\quad\Rightarrow\quad \tan x \:=\:\frac{4}{3}$
Since $\displaystyle \tan x \;=\:\frac{4}{3} \;=\:\frac{opp}{adj}$, we have: .$\displaystyle opp = 4,\;adj = 3\quad\Rightarrow\quad hyp = 5$
. . Hence: .$\displaystyle \sin x = \frac{4}{5},\;\cos x = \frac{3}{5}$
Therefore: .$\displaystyle D \;=\;\frac{2000}{4\sin x + 3\cos x} \;=\;\frac{2000}{4(\frac{4}{5}) + 3(\frac{3}{5})} \;=\;\boxed{400\text{ m}}$ (b)