Q:

How to solve this? (Argument of the complex number in complex plane)

Let the $z \in C$ s.t. $|z-10i|= 6 $
$\newcommand{\Arg}{\operatorname{Arg}}$
Say the $\theta = \Arg(z)$
Find the maximum and minimum value of the $8\sin \theta + 6\cos \theta$

My trial)
Trying to solve by picture and graph, I considered $8i\sin \theta + 6\cos \theta$ instead of the $8\sin \theta + 6\cos \theta$
Then, locus of the  $8i \sin \theta + 6\cos \theta$ is the tangent line of the  circle $|z-10i|= 6 $
(Surely   $ {-4\over 3} \leq \tan \theta \leq {4 \over 3}  $ )
So the question is I can't figure out the next step.
P.s.)
The answer sheet said "it is clear that when the case $\cos \theta = \pm{3 \over 5}  $ is maximum and minimum"
But this word totally unclear for me. Why does that case have a max or min value?

A:

Let $z=r(\cos\theta+i\sin\theta).$
Thus, $$r^2\cos^2\theta+(r\sin\theta-10)^2=36$$ or
$$\sin\theta=\frac{r^2+64}{20r}\geq\frac{2\sqrt{64r^2}}{20r}=\frac{4}{5},$$ which gives
$$\arcsin\frac{4}{5}\leq\theta\leq\pi-\arcsin\frac{4}{5}.$$
Also, write $$8\sin\theta+6\cos\theta=10\sin\left(\theta+\arccos\frac{4}{5}\right).$$
Can you end it now? 
Actually, I assumed that $0\leq\arg{z}<2\pi.$
I got $$2.8\leq8\sin\theta+6\cos\theta\leq10.$$
I used the following reasoning.
By C-S $$8\sin\theta+6\cos\theta\leq\sqrt{8^2+6^2)(\sin^2\theta+\cos^2\theta)}=10.$$
The equality occurs for $(\sin\theta,\cos\theta)||(4,3)$, which gives $\theta=\arctan\frac{4}{3}\in\left[\arcsin\frac{4}{5},\pi-\arcsin\frac{4}{5}\right]$, which says that we got a maximal value.
Also, $$f(x)=8\sin{x}+6\cos{x}=10\sin\left(x+\arccos\frac{4}{5}\right)$$ is a concave function on $\left[\arcsin\frac{4}{5},\pi-\arcsin\frac{4}{5}\right],$ which says
$$\min\limits_{\left[\arcsin\frac{4}{5},\pi-\arcsin\frac{4}{5}\right]}f=\min\left\{f\left(\arcsin\frac{4}{5}\right),f\left(\pi-\arcsin\frac{4}{5}\right)\right\}=f\left(\pi-\arcsin\frac{4}{5}\right)=2.8.$$

