Difference between revisions of "2008 AMC 10B Problems/Problem 14"
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− | Triangle <math>OAB</math> has <math>O=(0,0)</math>, <math>B=(5,0)</math>, and <math>A</math> in the first quadrant. In addition, <math>\angle ABO=90^\circ</math> and <math>\angle AOB=30^\circ</math>. Suppose that <math>OA</math> is rotated <math>90^\circ</math> counterclockwise about <math>O</math>. What are the coordinates of the image of <math>A</math>? | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Triangle <math>OAB</math> has <math>O=(0,0)</math>, <math>B=(5,0)</math>, and <math>A</math> in the first quadrant. In addition, <math>\angle ABO=90^\circ</math> and <math>\angle AOB=30^\circ</math>. Suppose that <math>OA</math> is rotated <math>90^\circ</math> counterclockwise about <math>O</math>. What are the coordinates of the image of <math>A</math>? <!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
<math> | <math> |
Revision as of 18:54, 27 March 2015
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
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
.
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.