Trigonometry question on R cos (theta - alpha). http://d.pr/L96v
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?
Edit: I mean to write, the answer to my first answer was √cos(Θ+14)
[r\sin(x+\theta)\] where \[r=\sqrt{a^2+b^2}, \tan(\theta)=\frac{a}{b}\]if i recall correctly
oops you want cosine, sorry
same idea but this time \[r\cos(x - \theta)\] same r but \[\tan(\theta)=\frac{b}{a}\]
so we need \[\theta\] in degrees i take it, yes?
yes
i get \[\theta = -14.4\] giving \[\sqrt{17}\cos(x+14.4)\]
\[\sqrt{17}\cos(x+14.4)=2\] \[\cos(x+14.4)=\frac{2}{\sqrt{17}}\] \[x=\cos^{-1}(\frac{2}{\sqrt{17}})-14.4\]
between -180 and 180 degrees.
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