How to find the surface area of a spherical triangle?

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SUMMARY

The discussion focuses on calculating the surface area of a spherical triangle with a radius of 1 and angles A=120°, B=73.89°, C=56.31°, a=115.66°, b=90°, and c=60°. The correct formula to use is (ABC) = (A + B + C - π) r², which yields a surface area of 1.2254. A critical point raised is the necessity of converting angles from degrees to radians to ensure accurate calculations.

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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?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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