2012 AMC 10B Problems/Problem 21

Revision as of 18:28, 12 March 2012 by Dongha (talk | contribs)

When you see a and 2a, you could think of 30-60-90 triangles. Since all of the other's lengths are a, you could think that b is root3a. Drawing the points out, it is possible to have a diagram where b=root3a. So, b=root3a, so B:A=root3a $\textbf{(A)}\ sqrt {3}\qquad\textbf

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