Difference between revisions of "2008 AMC 10B Problems/Problem 14"
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This means that the coordinates of <math>A</math> are <math>\left(5,\frac{5\sqrt 3}3\right)</math>. | This means that the coordinates of <math>A</math> are <math>\left(5,\frac{5\sqrt 3}3\right)</math>. | ||
− | After we rotate <math>OA</math> <math>90^\circ</math> counterclockwise about <math>O</math>, it will | + | After we rotate <math>OA</math> <math>90^\circ</math> counterclockwise about <math>O</math>, it will be in the second quadrant and have the coordinates |
<math>\boxed{ \left( -\frac{5\sqrt 3}3, 5\right) }</math>. | <math>\boxed{ \left( -\frac{5\sqrt 3}3, 5\right) }</math>. | ||
So the answer is <math>\boxed{\text{B}}</math>. | So the answer is <math>\boxed{\text{B}}</math>. |
Revision as of 12:59, 7 June 2021
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 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
.
So the answer is
.
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.