2010 AMC 12A Problems/Problem 8
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 .
Solution 2(Trig and Angle Chasing)
Let . Let . Because is equilateral, we get , so . Because is equilateral, we get . Angles and are vertical, so . By triangle , we have , and because of line , we have . Because Of line , we have , and by line , we have . By quadrilateral , we have .
By the Law of Sines, we have \sin(60-x)=\frac{\sin(60+x)}{2}\implies 2\sin(60-x)=\sin(60+x)\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)2(\sin(60)\cos(-x)+\cos(60)\sin(-x))=\sin(60)\cos(x)+\cos(60)\sin(x)2\sin(60)\cos(x)-2\cos(60)\sin(x)=\sin(60)\cos(x)+\cos(60)\sin(x) \implies \sin(60)\cos(x)=3\cos(60)\sin(x)\sin(60)=\frac{\sqrt{3}}{2}\cos(60)=\frac{1}{2}\frac{\sqrt{3}}{2}\cos(x)=\frac{3}{2}\sin(x)\implies . The only value of that satisfies (because is an angle of the triangle)x=30^{\circ}\angle ACB60+x90$. The answer is (C)
-vsamc
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.