Find the sines of an isosceles triangle when given its area and height.
For example, when A=60 and h=5, the result is [0.3752 0.3752 0.6927].
This problem has been polished and additional test cases were added.
I believe all of the test solutions of this problem are wrong
I believe that this problem is fixed now.
I think there might still be some issues with the test suite, Grant. Currently, only test suites 1 and 3 have sum(arcsind(y_correct))=180 within a degree. For test suite 3, you need to take 180-arcsind(y_correct(3)) to get the sum to equal 180, but the sines of those two angles are still the same.
Please let me know if I am missing something - If you take the arcsin for the current answers for problem 2 (y_correct = [0.7174 0.7174 0.8608]), you get angles in degrees of [49.84 49.84 59.41], which does not add up to 180 degrees. If you take 180-59.41 to keep the sine of angle #3 consistent, you end up with a sum greater than 180 degrees. Since this example is two 3-4-5 right triangles joined together at the 4-side (since height=4 and Area=12), I think the answer for case 2 should be [0.8 0.8 0.96].
Yes, the test solutions are incorrect. If the height of the triangle is h, and if the sides are s, then the sine of the angle at the base is h/s. The test solutions have all calculated this as sin(h/s).
@William, Great catch! Thanks a lot. :)
Don't waste your time on this one, the tests are all wrong. All posted solutions use if/else or won't work (they might have worked once, before the tests were changed). I've tried using the sine definition, the opposite height, law of sines, etc (all of them agree at a solution that's different from the test cases). And if the tests aren't wrong, then the problem description needs to be fixed...
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