## adnanchowdhury Group Title Trigonometry question on R cos (theta - alpha). http://d.pr/L96v 2 years ago 2 years ago

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)

3. satellite73 Group Title

[r\sin(x+\theta)\] where $r=\sqrt{a^2+b^2}, \tan(\theta)=\frac{a}{b}$if i recall correctly

4. satellite73 Group Title

oops you want cosine, sorry

5. satellite73 Group Title

same idea but this time $r\cos(x - \theta)$ same r but $\tan(\theta)=\frac{b}{a}$

6. satellite73 Group Title

so we need $\theta$ in degrees i take it, yes?

yes

8. satellite73 Group Title

i get $\theta = -14.4$ giving $\sqrt{17}\cos(x+14.4)$

9. satellite73 Group Title

$\sqrt{17}\cos(x+14.4)=2$ $\cos(x+14.4)=\frac{2}{\sqrt{17}}$ $x=\cos^{-1}(\frac{2}{\sqrt{17}})-14.4$