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Referring to TriangleIdentity by 伍柒貳 a while ago, considering $\bigtriangleup$ ABC, it is proved that:

$$\sin^2A \equiv \cos^2B + \cos^2C + 2 \cos A\cos B\cos C (1*) $$

I want to take angle $A = 0$ to look at the angle relationship of parallel lines so created.

From the above $ \cos B + \cos C = 0 $ gives two solutions as :

$$ 2 \cos( \frac {B+C}{2} ) \cos( \frac{B-C}{2} ) = 0 \, (2*) $$

$$ B + C = \pi ; B - C = \pi; (3*)$$

With $ A + B + C = \pi $ the first gives Euclid's proposition ( Two parallel lines cut by any transversal, two angle sum on one side equals $ \pi$) as expected.

But the second part of (3*) gives

$$ B = \pi - A/2 , \, C = -A/2 $$

The second situation has me stumped. Is the Sine & Cos Triangle Law someway flawed? Is there a seed in it of hyperbolic geometry? Please comment on any related aspect. Regards

Community
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Narasimham
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1 Answers1

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There are three "triangle"s with $\angle A = 0 $ and $AB $ parallel $AC$

  • a "triangle" with $\angle A = 0 $, $\angle B = 0 $ , $\angle C = \pi $

  • a "triangle" with $\angle A = 0 $, $\angle B = \frac{\pi}{2} $ , $\angle C = \frac{\pi}{2} $

  • a "triangle" with $\angle A = 0 $, $\angle B = \pi $ , $\angle C = 0 $

Willemien
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  • Not necessarily. Consider the family of triangles $\triangle ABC$ with, say, $|\overline{AB}|$ of fixed length, and $\angle C = k;\angle B$ for some arbitrary (but positive) $k$. As $\angle A$ shrinks to $0$, vertices $B$ and $C$ certainly converge, but arguably the limiting values of $\angle B$ and $\angle C$ are $\pi/(k+1)$ and $k\pi/(k+1)$, respectively. Thus, when $B$ and $C$ coincide in a flat $\triangle ABC$ (your middle case), their angles can be taken as any supplementary pair whatsoever; without a limiting context, the best we can say is that those angles are "indeterminate". – Blue Jan 17 '15 at 13:29
  • @ Willemein: $\angle A = 0 $, $\angle B = \alpha $ , $\angle C = \pi- \alpha $ for arbitrary $\alpha$. – Narasimham Jan 17 '15 at 15:41
  • yes you are right, now I am wondering g where the A comes from in "But the second part of (3*) gives $ B=\pi − \frac{A}{2},C=\frac{−A}{2} $ " just a new variable with an old name? but then what is the problem to start with? – Willemien Jan 17 '15 at 16:44