We consider a sphere with a radius of 4000 metres. We start at point A and travel on a spherical line segment to point B, turn 60$^{\circ}$ to our left then travel on a spherical line segment to point C, turn 120$^{\circ}$ to our left then travel on a spherical line segment for 350 metres, taking you back to point A. We do this without travelling more than 2000 metres. How far apart are points A and B?
I've tried using the spherical sine law to find the desired length, but my answer produced contradictions so I'm lost with how to proceed.
Best Answer
Suppose $a,b,c$ are the arcs of the given spherical triangle $ABC$ and $R=4000$ and $b=350$. As you did mentioned in your question From the law of Sinus we do have: $$\frac{\sin (b/R)}{\sin B}=\frac{\sin (c/R)}{\sin C}$$ , which gives $\sin (c/R)=\frac{\sin (2 \pi/3)}{\sin (\pi/3)}\sin(b/R)=\sin(b/R)\implies c=b=350$ or $c=\pi R-b$ (the last one is excluded as it should be $a+b+c<2000$). Now we check if the total perimeter of the triangle is less than 2,000. From the second law of Cosines of spherical triangles we have:$$\cos (a/R)=\frac {3}{4} \cos A -\frac{1}{4}$$.
$$\cos A=\frac {3}{4} \cos (a/R) +\frac{1}{4}$$.
From the above system we get the solution:$\cos (a/R) =\frac{-5}{14} \implies a/R=1.936\implies a=1.936 R>4,000$
Of course, the last result does not comply with the restriction that $a+b+c\le 2000$.