Difference between revisions of "2014 AMC 10A Problems/Problem 22"
m (→Solution (Trigonmetry)) |
Math101010 (talk | contribs) (→Solution (Trigonometry)) |
||
Line 7: | Line 7: | ||
==Solution (Trigonometry)== | ==Solution (Trigonometry)== | ||
− | Note that <math>\tan 15^\circ=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3</math>. (If you do not know the tangent half-angle formula, it is <math>\frac{1-\cos \theta}{\sin \theta}</math>). Therefore, we have <math>DE=10\sqrt 3</math>. Since <math>ADE</math> is a <math>30-60-90</math> triangle, <math>AE=2 \cdot AD=2 \cdot 10=\boxed{\textbf{(E)} \: 20}</math> | + | Note that <math>\tan 15^\circ=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3</math>. (If you do not know the tangent half-angle formula, it is <math>\tan \frac{\theta}2= \frac{1-\cos \theta}{\sin \theta}</math>). Therefore, we have <math>DE=10\sqrt 3</math>. Since <math>ADE</math> is a <math>30-60-90</math> triangle, <math>AE=2 \cdot AD=2 \cdot 10=\boxed{\textbf{(E)} \: 20}</math> |
==Solution 2 (non-trig)== | ==Solution 2 (non-trig)== |
Revision as of 10:56, 24 January 2016
Problem
In rectangle , and . Let be a point on such that . What is ?
Solution (Trigonometry)
Note that . (If you do not know the tangent half-angle formula, it is ). Therefore, we have . Since is a triangle,
Solution 2 (non-trig)
Let be a point on line such that points and are distinct and that . By the angle bisector theorem, . Since is a right triangle, and . Additionally, Now, substituting in the obtained values, we get and . Substituting the first equation into the second yields , so . Because is a triangle, .
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.