# Did I apply the Ceva's theorem correctly to this problem?

I need to confirm the following solution. I'm making a mistake somewhere. But I can't find the error.

I apply the trigonometric form of the Ceva's theorem:

$$\frac{\sin \angle 3}{\sin \angle 4}× \frac{\sin \angle 1}{\sin \angle 2}×\frac{\sin \angle 5}{\sin \angle 6} =1$$

$$\frac{\sin \angle 3}{\sin \angle 4}=\frac {\sin \angle 2 ×\sin \angle 6}{\sin \angle 1 ×\sin \angle 5}$$

$$\angle 3+\angle 4=\phi, \angle 1=\alpha, \angle 2=\frac{180°-\phi}{2}-\alpha ,\angle 3=x, \angle 4=\phi-x , \angle 5= \frac{180°-\phi}{2}-θ, \angle 6=θ$$

$$\frac {\sin x}{\sin (\phi -x)}=\frac{\sin \beta ×\sin θ }{\sin \alpha ×\sin \gamma}$$

$$\tan x=\frac{\frac{\sin \beta ×\sin θ }{\sin \alpha ×\sin \gamma}×\sin \phi}{1+\frac{\sin \beta ×\sin θ }{\sin \alpha ×\sin \gamma}×\cos\phi}$$

$$\tan x=\frac{\sin \beta ×\sin θ×\sin \phi}{\sin \alpha × \sin \gamma+\sin \beta ×\sin θ×\cos\phi}$$

$$x= \arctan \frac{\sin \beta ×\sin θ×\sin \phi}{\sin \alpha × \sin \gamma+\sin \beta ×\sin θ×\cos\phi}$$

$$\angle OBC=\phi-\arctan \frac{\sin \beta ×\sin θ×\sin \phi}{\sin \alpha × \sin \gamma+\sin \beta ×\sin θ×\cos\phi}$$

Finally, $$\alpha=\sqrt[3]{66,666°},\beta=6,667°-\sqrt[3]{66,666°}, \gamma=6,667°-\arctan 0,10666, θ=\arctan 0,10666$$

MathLab says that,

$$x≈4,345102733435...°, \angle OBC= \phi-x≈166,666°-4,345102733435...°=161,654897266...°$$

Where is the error in my solution? Please show me where I made a mistake. Unfortunately, I can not see.

Thank you.

• Computers are allowed in math contests? – GNUSupporter 8964民主女神 地下教會 Dec 9 '18 at 14:20
• @GNUSupporter8964民主女神地下教會 Of course no. I mean, my answer creates contradiction with this answer :math.stackexchange.com/a/3026556/548054 – Elementary Dec 9 '18 at 15:43
• @Beginner: You "know" the answer from the various solutions given to your previous question, so try this: At every stage of your analysis, substitute the known angle measures into your equation to double-check its correctness. When an equation breaks, you'll know where the problem (or the first problem) lies. – Blue Dec 11 '18 at 22:24
• @Beginner, Can you clear me one thing, what do you mean by the comma separations - are they decimal points? – Awe Kumar Jha Dec 18 '18 at 7:28

The only problem with your symbolic manipulations I see is that $$\tan x<0$$, so $$x$$ is actually $$\pi + \arctan \frac{\sin \beta ×\sin θ×\sin \phi}{\sin \alpha × \sin \gamma+\sin \beta ×\sin θ×\cos\phi}$$ According to my computations, this gives the same result as the answer you refer to.
However, there seems to be a bigger problem with your numerical computations (it should have given you a negative value for $$x$$), but I'm not familiar with MathLab, so I cannot say for certain what the problem is. Are you sure you're not mixing degrees and radians together, though? In your screenshot, there is a degree sign at the value 166.666, but not at the other angles, so maybe MathLab interprets the other angle values as radians?