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 (-- triangle), we can use the proportion to find .
We find that
That means the coordinates of are .
Rotate this triangle counterclockwise around , and you will find that will end up in the second quadrant with the coordinates .
Note: To better visualize this, one can sketch a diagram.
Solution 2
As and is 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 be in the second quadrant and have the coordinates .
See also
2008 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.