Difference between revisions of "2010 AMC 12A Problems/Problem 8"
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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>. | 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>. | ||
+ | == Solution 2(Trig and Angle Chasing) == | ||
+ | Let <math>AB=2a, AC=a</math>. Let <math>\angle BAE=\angle ACD=x</math>. Because <math>\triangle CFE</math> is equilateral, we get <math>\angle FCE=60</math>, so <math>\angle ACB=60+x</math>. Because <math>\triangle CFE</math> is equilateral, we get <math>\angle CFE=60</math>. Angles <math>AFD</math> and <math>CFE</math> are vertical, so <math>\angle AFD=60</math>. By triangle <math>ADF</math>, we have <math>\angle ADF=120-x</math>, and because of line <math>AB</math>, we have <math>\angle BDC=60+x</math>. Because Of line <math>BC</math>, we have <math>\angle AEB=120</math>, and by line <math>CD</math>, we have <math>\angle DFE=120</math>. By quadrilateral <math>BDFE</math>, we have <math>\angle ABC=60-x</math>. | ||
+ | By the Law of Sines, we have <math>\frac{\sin(60-x)}{a}=\frac{\sin(60+x)}{2a}\implies </math>\sin(60-x)=\frac{\sin(60+x)}{2}\implies 2\sin(60-x)=\sin(60+x)<math>. By the sine addition formula(which states </math>\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)<math> by the way), we have </math>2(\sin(60)\cos(-x)+\cos(60)\sin(-x))=\sin(60)\cos(x)+\cos(60)\sin(x)<math>. Because cosine is an even function, and sine is an odd function, we have </math>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)<math>. We know that </math>\sin(60)=\frac{\sqrt{3}}{2}<math>, and </math>\cos(60)=\frac{1}{2}<math>, hence </math>\frac{\sqrt{3}}{2}\cos(x)=\frac{3}{2}\sin(x)\implies <math>\tan(x)=\frac{\sqrt(3)}{3}</math>. The only value of <math>x</math> that satisfies <math>60+x<180</math>(because <math>60+x</math> is an angle of the triangle)<math> is </math>x=30^{\circ}<math>. We seek to find </math>\angle ACB<math>, which as we found before is </math>60+x<math>, which is </math>90$. The answer is (C) | ||
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+ | -vsamc | ||
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=7|num-a=9|ab=A}} | {{AMC12 box|year=2010|num-b=7|num-a=9|ab=A}} |
Revision as of 16:39, 10 March 2020
Problem
Triangle has . Let and be on and , respectively, such that . Let be the intersection of segments and , and suppose that is equilateral. What is ?
Solution
Let .
Since , triangle is a triangle, so .
Solution 2(Trig and Angle Chasing)
Let . Let . Because is equilateral, we get , so . Because is equilateral, we get . Angles and are vertical, so . By triangle , we have , and because of line , we have . Because Of line , we have , and by line , we have . By quadrilateral , we have .
By the Law of Sines, we have \sin(60-x)=\frac{\sin(60+x)}{2}\implies 2\sin(60-x)=\sin(60+x)\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)2(\sin(60)\cos(-x)+\cos(60)\sin(-x))=\sin(60)\cos(x)+\cos(60)\sin(x)2\sin(60)\cos(x)-2\cos(60)\sin(x)=\sin(60)\cos(x)+\cos(60)\sin(x) \implies \sin(60)\cos(x)=3\cos(60)\sin(x)\sin(60)=\frac{\sqrt{3}}{2}\cos(60)=\frac{1}{2}\frac{\sqrt{3}}{2}\cos(x)=\frac{3}{2}\sin(x)\implies . The only value of that satisfies (because is an angle of the triangle)x=30^{\circ}\angle ACB60+x90$. 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 |
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