# Angles determining a point interior to a triangle

Given a triangle $$ABC$$, a point $$P$$ interior to the triangle can be determined by two angles, for example the angle $$\alpha = \angle PAC$$ and the angle $$\beta = \angle PBA$$ (See diagram below).

In this case, once $$\alpha$$ and $$\beta$$ are chosen, the third similarly defined angle $$\gamma = \angle PCB$$ is fixed. This question is about how $$\gamma$$ depends on (the known) $$\alpha$$ and $$\beta$$.

Applying the sine rule to the three triangles meeting at $$P$$, I was able to find a formula

$$\cot \gamma = \cot C + \frac{\sin \alpha \, \sin \beta}{\sin (A-\alpha) \, \sin(B-\beta) \, \sin C} \; \cdot$$

Blindly applying trig formulae in this way leads to what looks like a quite complex expression and it is not directly obvious how it relates to what appears to be a simple geometric relationship.

Does anyone know of a simpler way to represent $$\gamma$$ and/or a basic geometric intuition to relate $$\gamma$$ to $$\alpha$$ and $$\beta$$?

• See the trigonometric form of Ceva's Theorem. This gets you to the same less-than-pretty relation you found, but at least shows that the relation derives from an elegant starting place.
– Blue
Commented Apr 11, 2021 at 21:28
• @Blue Another "elegant stating place" can be obtained by using barycentric coordinates (see my solution). I wouldn't be astonished that it is equivalent to the interesting trigonometric form of Ceva's theorem you mention and that I didn't know. Commented Apr 11, 2021 at 23:08
• @JeanMarie: Your determinant form is pretty neat, too. :) ... For some reason, Trigonometric Ceva (and Menelaus) isn't as well-known as the classical form. (That's why I had to link to Cut-the-Knot instead of, say, Wikipedia.) I've found it to be quite convenient, though.
– Blue
Commented Apr 11, 2021 at 23:13

The equations of lines $$PA,PB,PC$$ in barycentric coordinates $$(a,b,c)$$ are resp.

$$\begin{cases}0a&+&\sin(A-\alpha)b&-&\sin(\alpha)c&=&0\\ -\sin \beta a &+& 0b &+&\sin(B-\beta)c&=&0\\ \sin(C-\gamma)a&-&\sin \gamma b &+& 0c&=&0 \end{cases}$$

Therefore, these lines having a common point $$P$$, we can write a quite symmetric relationship under the form of the determinant of their coefficients equal to $$0$$:

$$\begin{array}{|ccc|}0&\sin(A-\alpha)&-\sin(\alpha)\\ -\sin \beta &0&\sin(B-\beta)\\ \sin(C-\gamma)&-\sin \gamma&0 \end{array}=0$$

or :

$$\sin \alpha \sin \beta \sin \gamma = \sin (A-\alpha) \sin (B-\beta) \sin (C-\gamma)$$

which is equivalent to the trigonometric form of Ceva's formula given by @Blue.

• @Blue Another rich site for different forms of Ceva theorem (and geometry in general!) Commented Apr 12, 2021 at 10:23
• Thank you both, this gives a very nice setting. The product of sines equation, whether via Ceva's theorem or barycentric coordinates, is easily rearranged to what I had found. Commented Apr 12, 2021 at 18:14