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While going through my phone's old pictures, I stumbled on a problem someone had sent me a while back. I remember trying hard, giving up, and then forgetting about it.

triangle problem

I tried to see how I can solve it via trigonometry or properties of isosceles triangles, but to no avail.

How is this problem to be solved?

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  • $\begingroup$ What have you done so far? If you show what you have done, we will be able 1) of showing you how to get to the answer or 2) find an error in your reasonning. I think one way to start is to look at the interior angles of the triangles. Remember that the sum of the interior angles of a triangle is 180°. $\endgroup$ – Bernard Massé Jan 17 at 13:54
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This is straightforward using trig form of Ceva,

$$\sin \angle OAC\sin \angle OCB\sin \angle OBA = \sin \angle OAB\sin \angle OBC\sin \angle OCA$$

Substituting $\angle OBC = x$ and $\angle OBA = 100-x$ and solving $$x=80^{\circ}$$

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  • $\begingroup$ What does Ceva mean? $\endgroup$ – Bernard Massé Jan 17 at 13:55
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    $\begingroup$ @BernardMassé Ceva's theorem. Please see link. $\endgroup$ – cosmo5 Jan 17 at 13:56
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Here's a way to do it without Ceva.

From the diagram, we have $$ \begin{align} \angle POC&=30^\circ,\\ \angle AOC &= 150^\circ,\\ \angle B &=100^\circ,\\ \angle POC&=30^\circ. \end{align}$$ Since $\triangle CPO$ is isosceles, we may take $\overline{OP}=\overline{CP}=1.$ Applying the law of cosines to $\triangle CPO$ gives$$\overline{OC}^2=1+1-2\cos(120^\circ)\implies\overline{OC}=\sqrt3$$

Applying the law of sines in $\triangle ACO$ gives $$\frac{\overline{AO}}{\sqrt3}=\frac{\sin10^\circ}{\sin20^\circ}$$ so that $$\overline{AO}=\frac{\sqrt3}{2\cos10^\circ}$$

In $\triangle BOP,$ the law of sines gives$$\frac{\overline{OB}}{\overline{OP}}=\frac{\sin60^\circ}{\sin x},$$ or $$\overline{OB}=\frac{\sqrt3}{2\sin x}\tag1$$

In $\triangle BOA,$ the law of sines gives $$ \frac{\overline{OB}}{\overline{OA}} =\frac{\sin20^\circ}{\sin(100^\circ-x)}$$ so that $$\overline{OB}=\frac{\sqrt3}{2\cos10^\circ}\cdot\frac{2\sin10^\circ\cos10^\circ}{\sin(100^\circ-x)}=\frac{\sqrt3\sin10^\circ}{\sin{(100^\circ-x)}}\tag2$$

Combining $(1)$ and $(2)$ gives $$2\sin10^\circ\sin x=\sin(100^\circ-x)$$ and we see that $$\boxed{x=80^\circ}$$ since $\sin80^\circ=\cos10^\circ.$

I did it wrong half a dozen times before getting it right, so Ceva is better.

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