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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 $cos(A-B) + cos(B-C) + cos(C-A)$

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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 $x$.

Sri Krishna
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    You can surely think of a better title. – Mariano Suárez-Álvarez May 28 '13 at 18:02
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    Do you know the trigonometric form of Ceva's theorem? – Calvin Lin May 28 '13 at 18:22
  • You can work out directly all of the angles, except PCB and PBC , using the sum of angles in a triangle and the sum around a single point. (I am presuming you did this.) You will have a value for angle A , and should find that PCB and PBC are complementary. The Law of Sines should permit you to find sines for angles B and C in terms of a, b, and c, and the value of the sine for angle A. You can get expressions for the cosines of all three angles, and all of this can be used with the identity for the cosine of the difference of two angles. (Though I like CL's suggestion for using Ceva...) – colormegone May 28 '13 at 18:44
  • @calvin : nope never heard of it . – Sri Krishna May 29 '13 at 02:12
  • @recklessreckoner : i did everything clearly and the evaluation part becomes too nasty to handle. – Sri Krishna May 29 '13 at 02:15
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    @SriKrishna Click on the link in my comment to understand it. It doesn't seem to help in this question though. Sadly, I agree that the evaluation part is nasty. – Calvin Lin May 29 '13 at 02:46
  • Yes, it is pretty nasty; there doesn't seem to be anything to be done about that. Since the sides of the triangle are unspecified, the measure of $ \ \angle PCB \ $ (what you are calling $ \ x \ $ ) and $ \ \angle PBC \ $ , which is complementary, appear to be dependent on those sides, so will not have specific numerical values. We can say that the measure of $ \ \angle A \ $ is 50º , but the sines and cosines of $ \ \angle B \ \text{and} \ \angle C \ $ will only have expressions. (The fact that no one else seems to have answered this problem yet suggests that the result is untidy.) – colormegone May 29 '13 at 16:51
  • @CalvinLin I think we may have given up too quickly on Ceva -- thank you for the suggestion, since it looks like that's what this problem was after... – colormegone May 30 '13 at 20:05

1 Answers1

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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.)

colormegone
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  • It's debatable about whether a calculator is allowed for the problem; I normally assume that calculators are not allowed, unless otherwise stated (which happens rarely). You can't get an exact answer for the $\sum \cos(A-B)$ portion with a calculator. – Calvin Lin Jun 02 '13 at 20:57