2014 AMC 10A Problems/Problem 22
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 | |
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