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I found this geometry question in a math video I watched recently.

enter image description here

$\triangle ABC$ is an equilateral triangle and point O is a random point taken inside the $\triangle ABC$ such that,

$\angle OAB=x, \angle OBC=42, \angle OCB=54$

The question is to find the value of $\angle x$. The person in the video solved this by using trigonometric ratios and the Ceva's theorem.

The final answer is $\angle x = 48$

My approach:

I tried very much time to solve the problem I reflect point O as the line of reflection would be BC and using some algebra by calculating angle BOC nothing wasn't helpful for me to find the value of angle x

But unfortunately I don't like trigonometry and I don't know Ceva's theorem.

So anyone in this community could help me to solve the problem.

Thank you !

sillysillybag69
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2 Answers2

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I think the comment by mathlove points the answer in the right direction. The second proof of that proof (beginning at "Here is the second proof") uses Euclidean geometry (only?) and no trigonometry.

That proof starts by knowing angles OCA=6° (ABD in the proof) and OAC=2·OCA (called BAD in the proof). From now on in this paragraph, I use nomenclature of that proof. Being BAD the double of ABD is crucial for the next steps, which identify some isosceles triangles. Eventually, it is proved that angle FDA=6·ABD equals angle FCG=(R/3)+ABD, being R a Right angle, which is crucial to prove that segments CD and FG intersect at right angles. Then it follows that ACD=(R/3)-2·ABD. The point is: this proof only works if (R/3)+ABD=6·ABD, i.e., if ABD=R/15=6° (also, all angles are multiples of 6°).

Thus we have a proof, using only (?) Euclidean geometry and no trigonometry, that an equilateral triangle can be partitioned into three triangles with angles (6,12,162)°, (18,48,114)°, and (42,54,84)°. Now, if we are requested to find $x$ in the picture of the question, when we know that OBC on the left is 42° and OCB on the right is 54°, then previous partition shows that $x=48°$.

James Gordon
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Of course the question itself is trigonometry, so any answer will contain some form of trigonometry. I will try to answer using a Cartesian coordinate description of the problem. It will just use the functions ``tan" and ``arctan" and some obvious properties.

Let $B=(B_x,B_y)=(0,0)$ and $C=(C_x,C_y)=(1,0)$. This can be done without loss of generality.

We have $\frac{O_y}{O_x}=\tan(42^\circ)$ and $\frac{O_y}{1-O_x}=\tan(54^\circ)$. Dividing the first equation by the second implies $$\frac{1-O_x}{O_x}=\frac{\tan(42^\circ)}{\tan(54^\circ)}.$$ Therefore $$\frac{1}{O_x}-1=\frac{\tan(42^\circ)}{\tan(54^\circ)}.$$ Therefore we can calculate $O_x$ as $$O_x=1/\Big(\frac{\tan(42^\circ)}{\tan(54^\circ)}+1\Big)\approx 0.6045284633.$$ We can therefore also calculate $O_y$ as $$O_y=O_x\tan(42^\circ).$$ Clearly $A_x=0.5$. Also quite clearly $A_y/A_x=\tan(60^\circ)$. Therefore $$A_y=0.5\tan(60^\circ)\approx 0.8660254038.$$ We get $$\angle x=30^\circ+\arctan\Big(\frac{O_x-A_x}{A_y-O_y}\Big)=30^\circ+\arctan\Big(\frac{O_x-0.5}{1/(2\tan(30^\circ))-O_x\tan(42^\circ)}\Big)\approx 48.00000000^\circ.$$

There's your answer.

I think this Cartesian approach is good to understand as an alternative approach. However, I would like to argue the trigonometric approach is

  • Shorter
  • More elegant
  • More precise (I used a calculator, so I can only guess that $48^\circ$ is the precise answer. Sometimes the answer will be something like $\frac{1}{2}\sqrt{3}$ and this precise answer can be hard or impossible to extract from the decimal number on your calculator.)
Teun
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