Question 933855
In a triangle ABC, AC is 10, BC is 0.025 and AB is 9.978. And the angle between BCA is said to be 30 degree. I need to find the other two angles. I have used the cosine law of triangles and many other methods. For each method, i am getting a different solution for these two angles CAB and CBA. Can you help me in solving this?.
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Using the given sides, BCA is not 30 degs, it's ~ 88.8494 degs
Use the Law of Sines to find the other angles.
Angle B = 5.738 degs
Angle A = 85.413 degs