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vera_ewing
 one year ago
Math question
vera_ewing
 one year ago
Math question

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0hint: \[\sin \theta = \pm \sqrt {1  {{\left( {\cos \theta } \right)}^2}} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0hint: \[\large \begin{gathered} \sin \theta = \pm \sqrt {1  {{\left( {\cos \theta } \right)}^2}} = \pm \sqrt {1  \frac{{16}}{{49}}} = \pm \sqrt {\frac{{33}}{{49}}} = ...? \hfill \\ \hfill \\ \tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{{\left( { \pm \sqrt {\frac{{33}}{{49}}} } \right)}}{{  \frac{4}{7}}} = ...? \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0hint: \[ \pm \sqrt {\frac{{33}}{{49}}} = \pm \frac{{\sqrt {33} }}{7}\]

vera_ewing
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino Okay, so sin0 = (sqrt)33/7 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not to butt in but this is somewhat easier if you draw a triangle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436190352602:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there is picture of an angle whose cosine is \(\frac{4}{7}\) find the "opposite" side via pythagoras then you can take any trig ratio you like

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0since we have 2 square roots, then: \[\sin \theta = \pm \frac{{\sqrt {33} }}{7}\]

vera_ewing
 one year ago
Best ResponseYou've already chosen the best response.0How do we find tanθ Michele?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0it is simple, you have to compute this ratio: \[\large \tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{{\left( { \pm \frac{{\sqrt {33} }}{7}} \right)}}{{  \frac{4}{7}}} = \left( { \pm \frac{{\sqrt {33} }}{7}} \right) \times \left( {  \frac{7}{4}} \right) = ...?\]

vera_ewing
 one year ago
Best ResponseYou've already chosen the best response.07(sqrt)33/28 @Michele_Laino ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436190696248:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0memorize the trig ratios, look at the triangle and you can find any trig ratio you like

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for example \[\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}=\pm\frac{\sqrt{33}}{7}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}=?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there is really very very little work to solve these just find the missing side of the triangle and use the trig ratios you already know

vera_ewing
 one year ago
Best ResponseYou've already chosen the best response.0tanθ = (sqrt)33/4 right?
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