Hi All,

for a triangle ABC, angle B is 50 degrees. Side AB is twice that of AC. Find angle A.

This means that side AB= x and AC= 2x. Then I thought the only way I could find angle A is if i could find x and then use the Sine rule to find A. But I don't have side AB? Can I assume this is 3x...?

Using the Cosine rule equation: (2x)^2 = (x)^2 + (3x)^2 - 2(x)(3x) Cos50

I continue with this to solve for x...?

Thanks