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 $\beta$?

  • 4
    $\begingroup$ 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. $\endgroup$
    – Blue
    Commented Apr 11, 2021 at 21:28
  • 1
    $\begingroup$ @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. $\endgroup$
    – Jean Marie
    Commented Apr 11, 2021 at 23:08
  • $\begingroup$ @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. $\endgroup$
    – Blue
    Commented Apr 11, 2021 at 23:13

1 Answer 1


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.

  • $\begingroup$ @Blue Another rich site for different forms of Ceva theorem (and geometry in general!) $\endgroup$
    – Jean Marie
    Commented Apr 12, 2021 at 10:23
  • $\begingroup$ 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. $\endgroup$ Commented Apr 12, 2021 at 18:14

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