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anonymous
 one year ago
The point (2, 3) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of θ? Make sure to show all work.
anonymous
 one year ago
The point (2, 3) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of θ? Make sure to show all work.

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1dw:1438268412570:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So can you explain that? lol

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1it is the representation of your problem

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, we can write this: \[\Large OP = \sqrt {{2^2} + {3^2}} = ...\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I have applied the Theorem of Pitagora to the triangle OP1P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the sin tan and cos?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes, we have to compute OP first

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1hint: \[\Large OP = \sqrt {{2^2} + {3^2}} = \sqrt {4 + 9} = ...?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it just square root of 13

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no, since we have to apply these formulas: \[\Large \begin{gathered} \cos \theta = \frac{{O{P_1}}}{{OP}} = \frac{2}{{\sqrt {13} }} = ... \hfill \\ \hfill \\ \sin \theta = \frac{{P{P_1}}}{{OP}} = \frac{3}{{\sqrt {13} }} = ... \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1finally, by definition, we can write: \[\Large \tan \theta = \frac{{P{P_1}}}{{O{P_1}}} = ...\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sweet thanks for the help!
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