Is this a poorly worded question?

  • Thread starter rockyshephear
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In summary, the original problem involved finding the components of a vector with a length of 10 and an angle of p/6 with the x-axis. However, the given solution used radian measure and found p to be equal to pi radians, or 180 degrees. The resulting vector found was 5sqrt(3)i + 5j.
  • #1
rockyshephear
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Example found on a website...(there were no graphs or explanations as to p or r or q)
Reverse engineering the question from the answer, I found p to be 180.
--------------------------------------------------------------------------

A vector that has length 10 makes an angle of p/6 with the x-axis. Find its components.


Solution:

x = r cos q, y = r sin q

So that

x = (10)(/ 2), y = 10 (1/2) = 5

We can write the vector as

5 i + 5j
 
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  • #2
They are using radian measure, and they meant p = [itex]\pi[/itex] radians which is equivalent to 180 degrees. Radian measure is derived from the unit circle; the radian measure of an angle is given by the length of arc that the angle subtends on a unit circle centered at the vertex of the angle. this makes use of the derived fact that the circumference of a circle is known to be 2[itex]\pi[/itex], so that 360 degrees is 2[itex]\pi[/itex] radians. It is common to leave off the "radian" as a unit of measure since it is also defined as a dimensionless ratio of two lengths (the length of arc divided by the length of the radius of the circle).
 
Last edited:
  • #3
Thanks
 
  • #4
Solution:

x = r cos q, y = r sin q

So that

x = (10)(/ 2), y = 10 (1/2) = 5

We can write the vector as

5 i + 5j
Is it just me, or does that vector have neither the length nor the angle specified in the original problem?
 
  • #5
yeah. answer is [tex]5\sqrt{3}i + 5j[/tex]

[tex]\cos (\frac {\pi}{6}) = \frac{\sqrt3}{2}[/tex]
 

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