Quandaries and Queries


I found a geometry problem that reads as follows:In triangle ABC, <C = 3<A, side BC = 27 and side AB = 48.Find the length of of side AC. How do I solve it? I tried using the law of sin and got Sin 3X/Sin X = 48 / 27. After simplifying I get Sin3 X = -7/36. Am I even on the right track or is there an easier way to do this?


Hi Jason,

I agree with your method but I don't get the result that you got. You let the angle at A be X degrees and used the law of sines to produce

Sin(3X)/Sin(X) = 48 / 27

The expression I used for Sin (3X) is

Sin(3X) = 3 Cos2(X) Sin(X) - Sin3(X)

and hence

Sin(3X)/Sin(X) = 48 / 27

simplifies to

3 Cos2(X) - Sin2(X) = 48/27

Using the fact that Cos2(X) = 1 - Sin2(X) I then get

Sin(X) = Sqrt(11)/6



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