How to find the surface area of a spherical triangle?

prinsinn
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Hello

I have a spherical triangle with the radius 1, and I have tried so hard to find the surface area. I know that A=120°, b=90° and c=60°.

I could calculate that B=73.89° and C=56.31° and a=115.66°.

I think I should use the formula
(ABC) = (A + B + C - pi) r2

I always get the wrong answer but the correct one is 1.2254.

Can someone please tell me what to do, thanks.
 
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Go on Wikipedia and Look up some of Gausses work on Spherical Trig, it should have an analogous result to Euclidean Planes area of a triangle (half a b sine c).
 
Girard's Theorem states that the sum of the angles of a triangle drawn on a sphere is not 180 degrees but the sum differs from 180 degrees by the area of the triangle divided by the radius squared.
 
hello prinsinn! :smile:

(have a pi: π :wink:)
prinsinn said:
I know that A=120°, b=90° and c=60°.

I could calculate that B=73.89° and C=56.31° and a=115.66°.

I think I should use the formula
(ABC) = (A + B + C - pi) r2

I always get the wrong answer but the correct one is 1.2254.

well, your formula is correct, and using your A B and C i do get 1.2254 :confused:
 
Did you remember to convert the angles to radians?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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