Difference between revisions of "2008 AMC 10B Problems/Problem 14"
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Since <math>\angle ABO=90^\circ</math>, and <math>\angle AOB=30^\circ</math>, we know that this triangle is one of the [[Special Right Triangles]]. | Since <math>\angle ABO=90^\circ</math>, and <math>\angle AOB=30^\circ</math>, we know that this triangle is one of the [[Special Right Triangles]]. | ||
− | We also know that <math>B</math> is <math>(5,0)</math>, so <math>B</math> lies on the x-axis. Therefore, <math> | + | We also know that <math>B</math> is <math>(5,0)</math>, so <math>B</math> lies on the x-axis. Therefore, <math>OB = 5</math>. |
− | Then, since we know that this is a Special Right Triangle(30-60-90 triangle), we can use the proportion <cmath>\frac{5}{\sqrt 3}=\frac{ | + | Then, since we know that this is a Special Right Triangle (<math>30</math>-<math>60</math>-<math>90</math> triangle), we can use the proportion <cmath>\frac{5}{\sqrt 3}=\frac{AB}{1}</cmath> to find <math>AB</math>. |
We find that <cmath>AB=\frac{5\sqrt 3}{3}</cmath> | We find that <cmath>AB=\frac{5\sqrt 3}{3}</cmath> | ||
− | That means | + | That means the coordinates of <math>A</math> are <math>\left(5,\frac{5\sqrt 3}3\right)</math>. |
Rotate this triangle <math>90^\circ</math> counterclockwise around <math>O</math>, and you will find that <math>A</math> will end up in the second quadrant with the coordinates | Rotate this triangle <math>90^\circ</math> counterclockwise around <math>O</math>, and you will find that <math>A</math> will end up in the second quadrant with the coordinates | ||
− | <math>\boxed{ \left( -\frac{5\sqrt 3}3, 5\right) | + | <math>\boxed{ \left( -\frac{5\sqrt 3}3, 5\right)}</math>. |
+ | |||
+ | Note: To better visualize this, one can sketch a diagram. | ||
==Solution 2== | ==Solution 2== | ||
− | As <math>\angle ABO=90^\circ</math> and <math>A</math> in the first quadrant, we know that the <math>x</math> coordinate of <math>A</math> is <math>5</math>. We now need to pick a positive <math>y</math> coordinate for <math>A</math> so that we'll have <math>\angle AOB=30^\circ</math>. | + | As <math>\angle ABO=90^\circ</math> and <math>A</math> is in the first quadrant, we know that the <math>x</math> coordinate of <math>A</math> is <math>5</math>. We now need to pick a positive <math>y</math> coordinate for <math>A</math> so that we'll have <math>\angle AOB=30^\circ</math>. |
By the [[Pythagorean theorem]] we have <math>AO^2 = AB^2 + BO^2 = AB^2 + 25</math>. | By the [[Pythagorean theorem]] we have <math>AO^2 = AB^2 + BO^2 = AB^2 + 25</math>. | ||
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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> | + | 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>. | ||
− | |||
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=B|num-b=13|num-a=15}} | {{AMC10 box|year=2008|ab=B|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 04:45, 21 August 2023
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 |
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