anonymous
  • anonymous
Trigonometry question on R cos (theta - alpha). http://d.pr/L96v
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Second part of the question is what i need help with. My answer to the first part: \[\sqrt{17} \cos (\Theta + \alpha)\] Second part: My answers are: 46.9 and -46.9. But answers says: 46.9 and -75?
anonymous
  • anonymous
Edit: I mean to write, the answer to my first answer was √cos(Θ+14)
anonymous
  • anonymous
[r\sin(x+\theta)\] where \[r=\sqrt{a^2+b^2}, \tan(\theta)=\frac{a}{b}\]if i recall correctly

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anonymous
  • anonymous
oops you want cosine, sorry
anonymous
  • anonymous
same idea but this time \[r\cos(x - \theta)\] same r but \[\tan(\theta)=\frac{b}{a}\]
anonymous
  • anonymous
so we need \[\theta\] in degrees i take it, yes?
anonymous
  • anonymous
yes
anonymous
  • anonymous
i get \[\theta = -14.4\] giving \[\sqrt{17}\cos(x+14.4)\]
anonymous
  • anonymous
\[\sqrt{17}\cos(x+14.4)=2\] \[\cos(x+14.4)=\frac{2}{\sqrt{17}}\] \[x=\cos^{-1}(\frac{2}{\sqrt{17}})-14.4\]
anonymous
  • anonymous
between -180 and 180 degrees.
anonymous
  • anonymous
Mark scheme: http://d.pr/ubFq Where does the value -75 come from?

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