# 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?

• "And I find trigonometry boring." Yeah.... that's.... not going to cut it. – fleablood May 8 at 15:48

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 $$\angle ACB=90-15=75^{\circ}$$ 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 $$\angle AOB = 150^0$$, $$\angle ACB = 75^0$$. 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 $$\angle EAC=\frac{\angle BOX}2 =15^o$$

$$\Rightarrow \angle{ACB}=90-15=75^o$$