Triangle geometry problem without trigonometry

Question

I have a problem here solving this without trigonometry. We have an acute triangle. Is given the following. Find the $\angle$ BCO.

I tried different approaches. The continuation of CO intersects AB in point M, such that we get an isosceles triangle BMO. I have taken the mid-vertical of AB and found useful angles. I take P on AB such that the $\angle$ AOP = 48°. So I will get POS isosceles triangle with PO = OC and $\angle$ CPO = $\angle$ PCO = 6°. Let the intersection point of BO and CP be K. $\angle$ KOP = 18°. $\angle$ OKC = 24°. I have $\angle$ OCP, now I need to find $\angle$ BCK, to find x.

I tried other approaches also but still did not find the solution.

Answer 1

Construct the equilater triangle $\triangle BOD$ as shown above and produce $OC$ to $E \in BD$. Take $F\in CE$ in such a way that $AF\cong AO$.

1. $AB$ perpendicularly bisects $OD$, hence $\measuredangle AOD = 42^\circ$, and $\triangle AOD$ is isosceles.
2. Then also $\triangle DAF$ is isosceles, since $AF \cong AO \cong AD$.
3. Angle chasing yields $\measuredangle AFC = \measuredangle FAC = 72^\circ$, so that $\triangle AFC$ is isosceles, too.
4. We get $\measuredangle FAD = 60^\circ$, so by 2. $\triangle DAF$ is in fact equilateral.
5. The quadrilateral $ACFD$ is thus a kite, whose diagonal $DC$ bisects its internal angles. Therefore, using 3., $\measuredangle OCD = 18^\circ$.
6. Finally, since by angle chasing $\measuredangle EOD = 30^\circ$, $CE$ perpendicularly bisects $DB$; by 5., $$\measuredangle OCD = \boxed{\measuredangle OCB = 18^\circ}.$$

Answer 2

Not the answer requested by the OP, but I thought it would be helpful to share the trigonometric solution:

First, let $a = AO = CO, b = BO$ and $y = \angle OBC$.

• Adding the angles of $\triangle ABC$ yields $36^\circ + 36^\circ +48^\circ + 30^\circ + x + y = 180^\circ$, hence $y = 30^\circ-x$
• Applying the law of sines to $\triangle AOB$ yields $\frac{\sin 30^\circ}{\sin 48^\circ} = \frac ab$.
• Applying the law of sines to $\triangle BOC$ yields $\frac{\sin (30^\circ - x)}{\sin x} = \frac ab$.

Combining these gives $$\frac{\sin 30^\circ}{\sin 48^\circ} = \frac{\sin (30^\circ - x)}{\sin x}$$ Now $\sin (30^\circ - x) = \sin 30^\circ\cos x - \cos 30^\circ\sin x$ So $$\frac{\sin 30^\circ}{\sin 48^\circ} = \sin 30^\circ\cot x - \cos 30^\circ$$ $$\cot x = \frac 1{\sin 48^\circ} + \cot 30^\circ$$ which yields $x = 18^\circ$.