Analytical Geometry HW: 1.2.1, mPR & mRN Solutions

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SUMMARY

The discussion focuses on solving the problem from Analytical Geometry HW 1.2.1, specifically calculating the slopes mPR and mRN. The correct calculations are mPR = 1/2, yielding an angle of approximately 26.57 degrees using arctan(1/2), and mRN = -1/5 with a reference angle of 11.32 degrees. The total angle R is calculated as 37.88 degrees. However, there is confusion regarding the notation \hat{R}, which typically denotes a unit vector, leading to ambiguity in the problem's requirements.

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DERRAN
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Homework Statement



http://img23.imageshack.us/img23/8680/68366634.gif


1.2.1 only!

Homework Equations





The Attempt at a Solution


mPR=1/2
tan(1/2)=26.57

mRN= -1/5
ref angle=11.32

R=11.32+26.57
=37.88

is it correct?
 
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DERRAN said:

Homework Statement



http://img23.imageshack.us/img23/8680/68366634.gif


1.2.1 only!

Homework Equations





The Attempt at a Solution


mPR=1/2
tan(1/2)=26.57

mRN= -1/5
ref angle=11.32

R=11.32+26.57
=37.88

is it correct?
I don't think so. The problem asks for [tex]\hat{R}[/tex]. The caret (or "hat") is usually interpreted to mean a unit vector, but that's probably not what is meant here, since the magnitude of any unit vector is 1. I think that what is asked for is the magnitude of the vector from O to R. You seem to have calculated the measure of angle PRQ. Since it's not clear to me what [tex]\hat{R}[/tex] is intended to represent, it's unclear to me what this problem is asking you to do.

Your calculation of the slope of PR is correct, but this isn't:
tan(1/2)=26.57
You can say that arctan(1/2) = 26.57 degrees, or that tan-1(1/2) = 26.57 degrees, where both are approximations.
That
 
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