Mollweide's proof | Mathematics | Page 1

jovial vapor Jun 16, 2023, 3:10 PM

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Guys, what are alpha and beta angles?

rugged belfry Jun 16, 2023, 3:37 PM

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If you recall properties of isosceles triangle, 2 of the angles are equal

jovial vapor Jun 16, 2023, 3:46 PM

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yeah i got what beta is

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C/2

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since A+B+C=180

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and triangle ACM has an angle of A+B

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okey i also got alpha is (A+B)/2

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i need now what is gamma

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i know it is 90 + something

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something = angle of BAN

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but i need it in terms of A and B, cuz i wanna apply sin theorem between c/sin(C/2) = (a+b)/sin(90+something)

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nvm, it is (A-B)/2 (the something)

jovial vapor Jun 17, 2023, 2:55 AM

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gamma is 90 degrees

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and sorry i don't know if this is what you mean, do you constrain line AB to be tangent to the circle C? if so, then AB and AC are perpendicular. all tangent lines to a circle are perpendicular to that circle's radius at the point where the circle, the radius and the tangent line intersect.

stable trail Jun 17, 2023, 3:30 AM

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Oh, seems like you resolved this yourself? Good job!

jovial vapor Jun 17, 2023, 11:49 AM

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jovial vapor gamma is 90 degrees

no

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90º is the - - - - - line

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not gamma

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gamma is a bit more

jovial vapor Jun 17, 2023, 5:47 PM

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ah, i see. so it looks to me gamma is the angle bam?

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then under the constraint (?) that BA is tangent to the circle C, i can reason that BA and BC are perpendicular. so given that the angle BCA is identical to alpha, and that beta and alpha are complement angles (this is because the triangle NAM is a right-angled triangle with the right angle at A), i reason that the angle B is identical to beta, etc. that's what i got out here at least.

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