2011 AMC 10B Problems/Problem 18
Contents
Problem
Rectangle has and . Point is chosen on side so that . What is the degree measure of ?
Solution 1
It is given that . Since and are alternate interior angles and , . Use the Base Angle Theorem to show . We know that is a rectangle, so it follows that . We notice that is a triangle, and . If we let be the measure of then
Easier Way to Continue
After finding we can continue using trigonometry as follows.
We know that and so
It is obvious that and so
Solving, we have
~mathboy282
Solution 2 (with trig, not recommended)
Let . If we let , we have that , by the Pythagorean Theorem, and similarily, . Applying the law of cosine, we see that and YAY!!! We have two equations for two variables... that are relatively difficult to deal with. Well, we'll try to solve it. First of all, note that , so solving for in terms of , we get that . The equation now becomes
Simplifying, we get
Now, we apply the quartic formula to get
We can easily see that is an invalid solution. Thus, .
Finally, since , , where is any integer. Converting to degrees, we have that . Since , we have that .
~ilovepi3.14
Solution 3(Easier Trig)
We have . By the Pythagorean Theorem, , and thus , We have , or ~awsomek
Solution 4 (elimination)
Let . Thus, . Since all angles should be positive, , narrowing the options to D and E. Trying (option D), is a 30-60-90 triangle. , so it follows that . Since , , too. However, that would imply that is also a triangle, which would, in turn, imply that , since . We know that and but we know that and . is clearly not , so this is not possible. Thus, the answer must be . ~ Technodoggo
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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All AMC 10 Problems and Solutions |
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