 | | Geometry —Fredrik Glöckner
| | | | Suppose I have a triangle, in which I know the lengths of the three sides. I want to find the corresponding angles inside the triangles. Is there a quick and simple way to do that? I know that I can divide the triangle into two triangles with a 90° (...) (25 years ago, 28-Jul-99, to lugnet.off-topic.geek)
| | | | | | | | | | Re: Geometry —Don Heyse
| | | | | (...) Reaching way, way back... Can you use the law of sines? sin A / a = sin B / b = sin C / c Seems like there's a step missing though. Perhaps the law of cosines is better: a*a + b*b + c*c = -2bc*cos(A) Solve for A? Anyhow, good luck. Don (25 years ago, 28-Jul-99, to lugnet.off-topic.geek)
| | | | | | | | | | | | Re: Geometry —Fredrik Glöckner
| | | | | (...) With a, b and c being the length of the sides, and A being the angle of the edge just opposite of a? Why, that looks just perfect! Thanks! (I'm used to using A, B and C for the lengths and the corresponding lowercase letters for the angles, so (...) (25 years ago, 29-Jul-99, to lugnet.off-topic.geek)
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