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In a spherical triangle the angles at α, β and γ are π/5, π/3, π/2. Find the sum of the sides, we shall call the sides a,b,c

So I'm looking at the formulas and I see one of Napier's rule which might work here.

Just to make things easier:

$$A=α=π/5$$ $$B=β=π/3$$ $$C=γ=π/2$$

So I want to use Napier's circle

Circle

So we need to find a,b,c and here is my work:

-> To find c: $$\sin(\pi-c)=\tan(\pi-B)\tan(\pi-A)$$ $$\cos(c)=\cot(B)\cot(A)$$ $$\cos(c)=\frac{\cos(\pi/3)}{\sin(\pi/3)}\frac{\cos(\pi/5)}{\sin(\pi/5)}$$ $$ c=37.38° $$

-> To find b: $$\sin(\pi-A)=\tan(b)\cos(\pi-B)$$ $$\cos(A)=\tan(b)\sin(B)$$ $$\tan(b)=\frac{\cos(\pi/5)}{\sin(\pi/3)}$$ $$ b=43.05° $$

-> To find a: $$\sin(\pi-A)=\cos(a)\cos(\pi-B)$$ $$\cos(A)=\cos(a)\sin(B)$$ $$\cos(\pi/5)=\cos(a)\sin(\pi/3)$$ $$ a=20.9° $$

-> sum of sides: $$a+b+c=20.9+43.05+37.38=101.3°$$

If anyone can verify if I did the whole process right that would be great. Also, note that the sum of the sides is 101.3°. I know spherical triangles have to be between 180° and 540° but that is for the sum of the angles, are there any rules like that for the sum of spherical triangle sides?

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    $\begingroup$ TL;DR... But i know that $a + b + c < 2 \pi$. $\endgroup$ – aGer Mar 18 '15 at 17:33

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