/sci/ - Science & Math

>>9713273
I'm not that far in cosines/sinus stuff but isn't cos/sin applicable only for right angle triangle?

I think I kinda solved it though, had to draw radius anyway to prove it:

triangle ABO - equilateral, therefore /_ AOB = 180 - 2a, were "a" is the base angle, both are equal.

/_ AOB also a central angle that means it is equal to lesser arc AB. However, since MA (M is the tangent line, M was just cut off-sreen) is a tangent line, that means OA _|_ MA and /_ MAO = 90. Therefore /_ MAB = 90-a == 1/2 AB arc because AB = 180-2a
Because both /_ AOB and /_ ACB share the same lesser arc AB and /_ ACB being a circumferential angle it means /_ ACB = 1/2 /_ AOB or 1/2 AB

Thus, /_ MAB = /_ ACB = half of the arc it covers inside the angle tangent line Ma and chord AB form.

i think it works but something bugs me