i was writing an paper on solutions of triangles when i encountered this sum -
In a $\Delta$ ABC , P is an interior point such that $\angle PAB = 10^\circ$ , $\angle PBA = 20^\circ$ , $\angle PCA = 30^\circ$ , $\angle PAC = 40^\circ$ and a , b , c are the sides of the triangle. Find the angles of the triangle and hence evaluate $cos(A-B) + cos(B-C) + cos(C-A)$
i used sine rule for the triangles but am clueless when i proceed to calculate $\angle PBC $ or $\angle PCB $. I assumed $\angle PCB = x $. My goal is to find $x$.
Here is what the extent of what I think we can say numerically for this set of triangles:
We are given that
$$m (\angle PAB) = 10º \ , \ m (\angle PAC) = 40º \ \Rightarrow \ m (\angle A) = 50º \ ; $$
$$m (\angle PBA) = 10º \ \Rightarrow \ m (\angle APB) = 160º \ ; $$
$$m (\angle PCA) = 30º \ \Rightarrow \ m (\angle APC) = 110º \ ; $$
$$ \Rightarrow \ m (\angle BPC) = 90º \ . $$
If we designate $ \ m (\angle PCB) = x \ $ , then we have $ \ m (\angle PBC) = 90º - x \ ; $ we can also conclude that $ \ m (\angle B) = 100º - x \ \text{and} \ m (\angle C) = 30º + x \ . $
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EDIT: All right, I think Calvin Lin was right after all about using the "trig form" of Ceva's Theorem. We still want the numerical information from above. Applying the Theorem here gives us
$$\sin(\angle PCA) \cdot \sin(\angle PBC) \cdot \sin(\angle PAB) \ = \ \sin(\angle PCB) \cdot \sin(\angle PBA) \cdot \sin(\angle PAC) $$
$$\Rightarrow \ \sin 30º \cdot \sin(90º - x) \cdot \sin 10º \ = \ \sin x \cdot \sin 10º \cdot \sin 40º $$
$$ \Rightarrow \ \sin 30º \cdot \cos x \ = \ \sin x \cdot \sin 40º \ \Rightarrow \ \tan x \ = \ \frac{\sin 30º}{\sin 40º} \ \approx \ 0.7779 $$
$$\Rightarrow \ m (\angle PCB) \approx 37.9º \ , \ \ m (\angle PBC) = 52.1º $$
$$\Rightarrow \ m (\angle B) \approx 62.1º \ , \ \ m (\angle C) = 67.9º . $$
So, if I used this properly, the shape of the triangle is unique and the lengths of the sides don't enter into the answer for the question. It looks like that Theorem is what it was intended that you apply here. (I now share CutTheKnot's lament that this theorem is not more widely taught.)