2008 AMC 10B Problems/Problem 14
Contents
Problem
Triangle has , , and in the first quadrant. In addition, and . Suppose that is rotated counterclockwise about . What are the coordinates of the image of ?
Solution 1
Since , and , we know that this triangle is one of the Special Right Triangles.
We also know that is , so lies on the x-axis. Therefore, .
Then, since we know that this is a Special Right Triangle(30-60-90 triangle), we can use the proportion to find .
We find that
That means that the coordinates of are .
Rotate this triangle counterclockwise around , and you will find that will end up in the second quadrant with the coordinates .
Solution 2
As and in the first quadrant, we know that the coordinate of is . We now need to pick a positive coordinate for so that we'll have .
By the Pythagorean theorem we have .
By the definition of sine, we have , hence .
Substituting into the previous equation, we get , hence .
This means that the coordinates of are .
After we rotate counterclockwise about , it will get into the second quadrant and have the coordinates . So the answer is .
See also
2008 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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