% Bernd Berg jun 29 2018.
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\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2018 
Solutions} \medskip

\centerline{\bf Set 1} 

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\item{\bf 3. Sum of angles in a geodesic triangle on a sphere.} 
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In the notation of Fig.~8.4 the area of the sphere is $A_S=4\pi\,a^2$
and, consequently, for the half-sphere $A_{S/2}=2\pi\,a^2$. The area
of the triangle with $r=a\,\pi/2$ is then
$$ A_{\theta} = \frac{\theta}{2\pi}\,A_{S/2} = \theta\,a^2\,. $$ 
Setting now $A_{\theta}=A_S/n=4\pi\,a^2/n$ we get
$$ \theta = \frac{4\pi}{n}\,. $$
Each of the other two angles has the values $\pi/2$. So we get for
the sum of the angles in this triangle:
$$ \sum_{\rm triangle}{\rm angles} = \pi+\theta = \pi+\frac{4\pi}{n} 
   = \pi\,\left(1+\frac{4}{n}\right)\,.$$
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