2014 AMC 10A Problems/Problem 22
Contents
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 (Without Trigonometry)
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, .
Solution 3 (Trigonometry)
By Law of SinesThus,
We see that is a triangle, leaving
Solution 4 (Measuring)
If we draw rectangle and whip out a protractor, we can draw a perfect , almost perfectly degrees off of . Then we can draw , and use a ruler to measure it. We can clearly see that the is .
NOTE: this method is a last resort, and is pretty risky. Answer choice is also very close to , meaning that we wouldn't be 100% sure of our answer. However, If we measure the angles of , we can clearly see that it is a triangle, which verifies our answer of .
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.