2010 AMC 12A Problems/Problem 8
Triangle has . Let and be on and , respectively, such that . Let be the intersection of segments and , and suppose that is equilateral. What is ?
Since and the angle between the hypotenuse and the shorter side is , 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 . By the sine addition formula(which states by the way), we have . Because cosine is an even function, and sine is an odd function, we have . We know that , and , hence . The only value of that satisfies (because is an angle of the triangle) is . We seek to find , which as we found before is , which is . The answer is
Solution 3 (Similar Triangles)
Notice that and . Hence, triangle AEB is similar to triangle CFA. Since , , as triangle CFE is equilateral. Therefore, , and since , . Thus, the measure of equals to -HarryW
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