Difference between revisions of "2010 AMC 12A Problems/Problem 8"
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\angle BAC &= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}</cmath> | \angle BAC &= \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}</cmath> | ||
− | Since <math>\frac{AC}{AB} = \frac{1}{2}</math>, <math>\angle BCA = \boxed{90^\circ\ \textbf{(C)}}</math> | + | Since <math>\frac{AC}{AB} = \frac{1}{2}</math>, triangle <math>ABC</math> is a <math>30-60-90</math> triangle, so <math>\angle BCA = \boxed{90^\circ\ \textbf{(C)}}</math> |
== See also == | == See also == |
Revision as of 21:52, 1 February 2011
Problem
Triangle has . Let and be on and , respectively, such that . Let be the intersection of segments and , and suppose that is equilateral. What is ?
Solution
Let .
Since , triangle is a triangle, so
See also
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |