|
Title: How do you find the angle of A when B is 100 degrees, C is unknown, and it is not a right triangle? Post by: FreedomTrain on Jul 4, 2013 I think it is the law of cosines and I now that that sinC/A=sinA/C.
Sorry side AB=3 and side BC=6. Title: How do you find the angle of A when B is 100 degrees, C is unknown, and it is not a right triangle? Post by: twinzmomma on Jul 4, 2013 Whether to use the Law of Sines or Cosines will really depend on what other data are given.
Based on what you have posted, there is no sufficient information as to what formula needs to be used. Hope this helps. Title: How do you find the angle of A when B is 100 degrees, C is unknown, and it is not a right triangle? Post by: troy.clements on Jul 4, 2013 You need minimum of three elements to solve any triangle.
Title: How do you find the angle of A when B is 100 degrees, C is unknown, and it is not a right triangle? Post by: legna95023 on Jul 4, 2013 The one you show ("sinC/A = sinA/C") is none that I know of. SinA/a = SinC/c is the Sine law but that won't work here, at least at first.. Cosine law will.
(AC)^2 = (AB)^2 + (BC)^2 - 2(AB)(BC)CosB (AC)^2 = (3)^2 + (6)^2 - 2(3)(6)(-0.1736) (AC)^2 = 9 + 36 + 6.2496 (AC)^2 = 51.2496 AC = 7.16 Now Sine law Sin100/7.16 = SinA/6 0.9848/7.16 = SinA/6........cross multiply 7.16SinA = 5.909 SinA = 0.8252 A = 55.6 degrees......answer Sin100/7.16 = SinC/3 7.16SinC = (0.9848)(3) SinC = 2.9544/7.16 C = 24.4 degrees.........answer Check 55.6 + 100 + 24.4 = 180 180 = 180 |