In $\triangle{ABC}$, $\angle ABC=45^ \circ$. $X$ is a point on $BC$ such that $BX=\frac{1}{3}BC$ and $\angle AXC=60^ \circ$. Find $\angle ACB$. Problem

In $\triangle{ABC}$, $\angle ABC=45^ \circ$. $X$ is a point on $BC$ such that $BX=\frac{1}{3}BC$ and $\angle AXC=60^ \circ$. Find $\angle ACB$.

The problem looks easy. Though I couldn't solve it in an efficient way. Finally I solved it using trigonometry.
Trig solution
Let $BX = a$ units, then $BC = 3a$ and $XC = 3a-a= 2a$ units. $\angle AXC =60^ \circ$ and $\angle ABC= 45^ \circ$, then $\angle BAX= 60^ \circ -45^ \circ = 15^ \circ$.
Applying sine rule in $\triangle ABX$,
$$\frac{BX}{\sin \angle BAX} = \frac{AX}{\sin \angle ABC}$$
$$\implies \frac{a}{\sin 15°} = \frac{AX}{\sin 45°} \tag{1}$$
In $∆AXC$ , let $\angle ACB = \theta$, then $\angle XAC = (120 - \theta)$ and by sine rule,
$$\frac{XC}{\sin \angle XAC} = \frac{AX}{\sin \angle ACB}$$
$$\implies \frac{2a}{\sin (120 - \theta)} = \frac{AX}{\sin \theta} \tag{2}$$
Dividing $(1)$ by $(2)$,
$$\frac{\sin (120-\theta)}{2\sin 15^ \circ}= \frac{\sin \theta}{\sin 45°}$$
$$\implies 2\sin 15°\cdot\sin \theta = \sin 45°\cdot\sin (120-\theta)$$
$$\implies \frac{\sqrt{3}–1}{\sqrt 2}.\sin \theta = \frac{1}{\sqrt 2}.(\sin120°.\cos \theta - \cos 120°.\sin \theta).$$
$$\implies (\sqrt{3}–1).\sin \theta = \frac{\sqrt 3}{2}.\cos \theta + \frac{1}{2}.\sin \theta$$
$$\implies \tan \theta= 2+\sqrt 2$$
$$\implies \theta=75^ \circ$$
Thus, $\angle ACB = 75°$.

This solution is impossible without knowing the values of $\sin 15^ \circ$ and $\tan 75^ \circ$. And I find trigonometry boring. So, can this problem be solved in some other ways?
 A: Drop a perpendicular from point $C$ on $AX$ and let the feet of this perpendicular be $D$.
$DX=\frac {1}{2}XC=BX$ and thereafter $BD=DC$ and $AD=BD$ by simple angle chasing. Thus $D$ is the circumcentre of $\triangle ABC$ and $\angle ACB=90-15=75^{\circ}$
A: 
Draw a perp from $C$ to $AX$. Now as $ \displaystyle \angle AXC = 60^0, OX = \frac{CX}{2} = BX$
As $\angle AXB = 120^0$, $\angle XOB = \angle OBX = 30^0$. Also, $\angle OCX = 30^0$.
So $OB = OC$.
Now $\angle ABO = \angle XAB = 15^0 \implies OA = OB$
$OA = OB = OC$ and hence $O$ is circumcenter of the $\triangle ABC$.
As $\angle AOB = 150^0$, $\angle ACB = 75^0$.
A: 
Considering figure we have:
$CH=\frac{3R}2$
where R is the radius of circucircle of triangle AXC. Also:
$BC=\frac{CH}{cos 45^o}=\frac{3R}{\sqrt 2}\rightarrow BC^2=\frac{9R^2}2\rightarrow \frac23 BC^2=3R^2$
$CX\cdot BC=\frac 23 BC\cdot BC=\frac 23 BC^2=CD^2\rightarrow CD^2=\frac 23 3R^2=2R^2$
In triangle ACX we have:
$\frac{AC}{sin 60^o}=2R\rightarrow \frac{AC^2}{\frac 34}=4R^2\rightarrow AC^2=3R^2=CD^2\rightarrow CD=CA$
That is CA is tangent to circumcircle of AXC and $\angle OAC=90^o$. Also $\angle XOA=90^o$ which means $XO\parallel AC$ and $\angle BOX=30^o$
Also $\angle EAC=\frac{\angle BOX}2 =15^o$
$\Rightarrow \angle{ACB}=90-15=75^o$
