Sum of internal angles of triangle in Poincare HP

bejou102 · **Joined:** Tue, 6 Sep 2011 04:11:20 UTC **Posts:** 19

I just went through the calculations for finding the internal angles of triangle ABC in the Poincare Half-Plane where A(1,1), B(1,2), and C(2,1). I found the sum of the internal angles to be approximately 176.65 degrees.

I'm just looking for confirmation that I did the calculations correctly. Thank you in advance.

Shadow · **Posted:** Fri, 10 Feb 2012 04:24:37 UTC

bejou102 wrote:

I just went through the calculations for finding the internal angles of triangle ABC in the Poincare Half-Plane where A(1,1), B(1,2), and C(2,1). I found the sum of the internal angles to be approximately 176.65 degrees.

I'm just looking for confirmation that I did the calculations correctly. Thank you in advance.

That looks right, just a hair short of $\pi$ is usually what you'll get with something like that.

bejou102 · **Joined:** Tue, 6 Sep 2011 04:11:20 UTC **Posts:** 19

Thank you. Do you know of any website that would have a calculator for working in the halfplane?

Shadow · **Posted:** Fri, 10 Feb 2012 04:45:25 UTC

bejou102 wrote:

Thank you. Do you know of any website that would have a calculator for working in the halfplane?

I doubt it.

outermeasure · **Posted:** Fri, 10 Feb 2012 10:34:44 UTC

bejou102 wrote:

I just went through the calculations for finding the internal angles of triangle ABC in the Poincare Half-Plane where A(1,1), B(1,2), and C(2,1). I found the sum of the internal angles to be approximately 176.65 degrees.

I'm just looking for confirmation that I did the calculations correctly. Thank you in advance.

That is definitely incorrect. The angular defect is $4\arctan(\frac{1}{2})-\frac{\pi}{2}=\arctan(\frac{7}{24})$ , since the angles at A,B,C (assuming you are using the upper half-plane rather than the right half-plane) are $\arctan(2),\arctan(2),\arctan(2)-\arctan(\frac{1}{2})$ respectively.

Shadow · **Posted:** Fri, 10 Feb 2012 18:17:20 UTC

outermeasure wrote:

bejou102 wrote:

I just went through the calculations for finding the internal angles of triangle ABC in the Poincare Half-Plane where A(1,1), B(1,2), and C(2,1). I found the sum of the internal angles to be approximately 176.65 degrees.

I'm just looking for confirmation that I did the calculations correctly. Thank you in advance.

That is definitely incorrect. The angular defect is $4\arctan(\frac{1}{2})-\frac{\pi}{2}=\arctan(\frac{7}{24})$ , since the angles at A,B,C (assuming you are using the upper half-plane rather than the right half-plane) are $\arctan(2),\arctan(2),\arctan(2)-\arctan(\frac{1}{2})$ respectively.

And that's why it's good to actually check, in spite of heuristics.

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Sum of internal angles of triangle in Poincare HP

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