Simpler solution to a geometry problem In a set of geometry problems, I got this one:

If in a triangle $ABC$ with segments $AB=8$, $BC=4$, and $3A+2B=180^{\circ}$, calculate the side $AC$

My solution was
Let $A=2\alpha$,$B=90^{\circ}-3\alpha$, where $\alpha<30$, then the second condition is always met.

So
$$tan(2\alpha)=\frac{cos(3\alpha)}{2-sen(3\alpha)}$$
$$\frac{2sen(\alpha)cos(\alpha)}{cos(2\alpha)}=\frac{cos(\alpha)(2cos(2\alpha)-1)}{2-sen(\alpha)(2cos(2\alpha)+1)}$$
$$\frac{2sen(\alpha)}{cos(2\alpha)}=\frac{2cos(2\alpha)-1}{2-sen(\alpha)(2cos(2\alpha)+1)}$$
$$4sen(\alpha)-2sen^2(2cos(2\alpha)+1)=2cos^2(2\alpha)-cos(2\alpha)$$
$$4sen(\alpha)+(cos(2\alpha)-1)(2cos(2\alpha)+1)=2cos^2(2\alpha)-cos(2\alpha)$$
$$4sen(\alpha)+2cos^2(2\alpha)-cos(\alpha)-1=2cos^2(2\alpha)-cos(2\alpha)$$
$$sen(\alpha)=\frac{1}{4}$$
We now construct the altitude to $BC$

And since $sin(\alpha)=\frac{1}{4}$ we set $AC=4k$, $CE=k$, $AE=\sqrt{15}k$
Then it follows from pythagoras that
$$(k+4)^2+15k^2=64$$
$$16k^2+8k-48=0$$
$$2k^2+k-6=0$$
$$(2k-3)(k+2)=0$$
Since $k$ is positive, $k=\frac{3}{2}\iff AC=4k=6$
But the $2,3,4$(from the sides $4,6,8$) pattern makes me think there is an easier way, so it makes me think I missed something obvious. Any hints are ideas are greatly appreciated.
 A: From A=2α, B=90-3α it follows that C=90+α since the sum of the angles are 180°
Now apply the Laws of Sine involving AB and BC, use an identity for sin(90+α) and you get a fairly straight forward trig equation with sinα and cosα from which you can find sinα = 1/4
Took me two lines...
PS, Great question!!!
A: 
We might also "drop a perpendicular" from vertex $ \ B \ $ to the extended line $ \ AC \ \ , \ $ the length of which we will call $ \ y \ \ . \ $  Using your notation for angle measure, we have $ \ y \ = \ 8 \sin (2 \alpha) \ = \ 4 \sin (90º - \alpha) \ = \ 4 \cos \alpha  \ \ . $  From this, we find
$$ 8 \ · \ 2 · \sin \alpha · \cos \alpha \ = \ 4 · \cos \alpha \ \ \Rightarrow \ \ \sin \alpha \ = \ \frac14 \ \ , \ $$ since $ \ \cos \alpha \ \neq \ 0 \ \ . $  (This is just a variant of imranfat's method, since the Law of Sines can be derived using altitudes of the triangle.)
The Law of Cosines then gives us
$$ x^2 \ \ = \ \ 8^2 \ + \ 4^2 \ - \ 2·8·4·\cos(90º - 3 \alpha) \ \ = \ \ 80 \ - \ 64 \sin (3 \alpha) \ \ , \ $$
with $ \ x \ $ being the length of side $ \ AC \ \ . $
From the "triple-angle" identity for sine, we obtain $$   \sin (3 \alpha) \ \ = \ \ 3·\sin \alpha \ - \ 4·\sin^3 \alpha \ \ = \ \ 3·\frac14 \ - \ 4·\left(\frac14 \right)^3 \ \ = \ \ \frac{11}{16} \ \ . \ $$
Thus, $$ x^2 \ \ = \ \  80 \ - \ 64·\frac{11}{16} \ \ = \ \ 80 \ - \ 44 \ \ = \ \ 36 \ \ \Rightarrow \ \ x \ = \ 6 \ \ . $$
[We can check the consistency of our calculations using our constructed right triangle: since $ \ \cos \alpha \ = \ \frac{\sqrt{15}}{4} \ \ , \ $ we have $ \ y \ = \ 4·\cos \alpha \ = \ \sqrt{15} \ \ , \ \  z \ = \ 4·\cos(90º - \alpha) \ = \ 4·\sin \alpha \ = \ 1 \ \ , \ $ and so
$$ y^2 \ + \ (x + z)^2 \ \ = \ \ (\sqrt{15})^2 \ + \ (6 + 1)^2 \ \ = \ \ 15 \ + \ 7^2 \ \ = \ \ 8^2 \ \ . \ ] $$
A: From $C$ draw the median $CM$, the line $r$ parallel to $AB$, and the line $s$  perpendicular to $CM$. From $A$ draw the line $t$ parallel to $CM$, that intersects $s$ in $N$. Also let $E$ a point in $s$ such that $\angle CAN \cong \angle NAE$. Finally, let $D$ be the intersection between $r$ and $AE$, and $P$ the intersection between $t$ and $r$.

*

*Using the hypothesis on $\measuredangle ABC$ and $\measuredangle BAC$, show that $\measuredangle CAN = \frac{\pi}4 - \frac{\measuredangle BAC}4$.

*Use the fact that $\measuredangle CAD = 2\cdot \measuredangle CAN$, and that $\measuredangle BAC = \measuredangle ACD$ to conclude that $\triangle CAD$ is isosceles.

*Show that Thales Theorem yields $AD \cong DE$.

*From 3. and Internal Bisector Theorem on $\triangle ACD$ conclude that $PD \cong \frac12 CP$.

*From the fact that $\overline{CP} = 4$ and 2. derive that $\overline{AC} = 6$.


