Because your teacher told you so, as a fait accompli !BigGlenntheHeavy said:\(\displaystyle Afterthought: \ Why \ is \ triangle \ ABD \ a \ right \ triangle?\)
Because your teacher told you so, as a fait accompli !BigGlenntheHeavy said:\(\displaystyle Afterthought: \ Why \ is \ triangle \ ABD \ a \ right \ triangle?\)
a = SIN(33), b = SIN(57), p = PIDenis said:Quite easy if we let the hypotenuse = 1 :
Ratio = SIN(33)SIN(57) / (PI x^2) where x = [SIN(57) + SIN^2(33) / SIN(57)] / 2
That's .409065723....
I'm trying to find the ratio of the area of this right triangle to the area of the semicircle.
The solution is supposed to be 0.409.
Denis said:a = SIN(33), b = SIN(57), p = PIDenis said:Quite easy if we let the hypotenuse = 1 :
Ratio = SIN(33)SIN(57) / (PI x^2) where x = [SIN(57) + SIN^2(33) / SIN(57)] / 2
That's .409065723....
Ratio = ab / (px^2) where x = (b + a^2 / b) / 2
I'm depressed: nobody seems impressed with my beautiful SINful solution
Well, sure more fun chasing pucks than getting dressed like a surgical doctor and bowling cricket balls:Subhotosh Khan said:No matter - how elegant - Who would ever want to join a hockey-puck chasing SINner??
markjay50 said:H..ALL !!!
I'm interested in seeing how you arrived at: hypotenuse = D cos(33°)...!!!
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