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Hi, I am new. Can someone help me with this trigonometry problem? I doing a couple of review questions my tutor gave me. Thank you. (;

 

 

 

Triangle PQR has vertices P(0, 1), Q(0, -4), and R(2, 5). Find the coordinates of R' to the nearest hundredth after rotating triangle PQR counterclockwise about the origin 45º.

Guest Oct 17, 2014

Best Answer 

 #6
avatar+85726 
+5

If we are not using rotational matrices, we can solve this using basic trig. Note that the angle between the x axis and a line drawn from the origin to the point (2, 5) is given by:

tan-1(5/2) = about 68.2°

And the length of this line (i.e., r) is given by √(22 + 52) = √(4 + 25) = √29

And rotating this line by 45° gives us ( 68.2° +  45°) = 113.2°

So the coordinates of the new point are given by [r * cos(113.2°), r * sin(113.2° ] = [√29*cos(113.2°), √29*sin(113.2° ] = 

(-2.12, 4.95)

And that's it......!!!!!

 

CPhill  Oct 17, 2014
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6+0 Answers

 #1
avatar+17721 
0

If you solve this by matrices, the rotational maxtrix for counter-clockwise is:  |  cosθ    -sinθ  |                                                                                                                                         |   sinθ     cosθ  |

(I know those bars don't look like matrix notation; but I don't know how to do better. Sorry.)

Rotational Matrix X Point Matrix:

|  cos45    -sin45  |    X    | 2 |

|  sin45     cos45  |          | 5 |

(Remember; those aren't absolute value bars; they're supposed to indicate 2 matrices.)

geno3141  Oct 17, 2014
 #2
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Im confused...Can you work it out for me? 

Guest Oct 17, 2014
 #3
avatar+17721 
+5

|   √2/2    -√2/2   |    x  | 2 |

|   V2/2       √2/2  |        | 5 |

Take the elements of the first row of the first matrix and multiply them, term by term, to the elements of the first (and, in this case, only) row of the second matrix:   (√2/2 x 2)  +  (-√2/2 x 5)  =  -3√2/2

Take the elements of the second row of the first matrix and multiply them, term by term, to the elements of the first (and, in this case, only) row of the second matrix:    (√2/2 x 2)  +  (√2/2 x 5)  =  7√2/2

Your final matrix is:  | -3√2/2 |

                               |  7√2/2 |

which states that R' is ( -3√2/2, 7√2/2 )

geno3141  Oct 17, 2014
 #4
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0

I still don't understand how to get the answer...Maybe I can give you the choices and that will help...

 

A) (4.95, 2.12)

 
 

B) (2.12, -2.12)

 
 

C) (2.12, -4.95)

 
 

D) (-2.12, 4.95)

 
 
Guest Oct 17, 2014
 #5
avatar+17721 
0

The answer is D:  -3√2/2 = -2.121      7√2/2 = 4.950

I'm sorry that I'm not communicating well with you; are you supposed to use matrices?

geno3141  Oct 17, 2014
 #6
avatar+85726 
+5
Best Answer

If we are not using rotational matrices, we can solve this using basic trig. Note that the angle between the x axis and a line drawn from the origin to the point (2, 5) is given by:

tan-1(5/2) = about 68.2°

And the length of this line (i.e., r) is given by √(22 + 52) = √(4 + 25) = √29

And rotating this line by 45° gives us ( 68.2° +  45°) = 113.2°

So the coordinates of the new point are given by [r * cos(113.2°), r * sin(113.2° ] = [√29*cos(113.2°), √29*sin(113.2° ] = 

(-2.12, 4.95)

And that's it......!!!!!

 

CPhill  Oct 17, 2014

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