Analytical Geometry HW: 1.2.1, mPR & mRN Solutions

AI Thread Summary
The discussion focuses on the homework problem 1.2.1, where the calculations for the slopes mPR and mRN are presented. The correct slope for mPR is confirmed as 1/2, but the tangent calculation is clarified as arctan(1/2) equating to 26.57 degrees, not tan(1/2). The value for mRN is noted as -1/5, with a reference angle of 11.32 degrees. There is confusion regarding the notation \hat{R}, with suggestions that it may refer to the magnitude of the vector from O to R rather than a unit vector. Overall, the calculations need clarification on the intended meaning of \hat{R} and the correct interpretation of the tangent function.
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 \hat{R}. 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 \hat{R} 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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