Respuesta :

tramserran

Answer:  19)  117°                    20) 53°

               21)  229°                   22) 119°

               23) 155°                    24) 323°

Step-by-step explanation:

Use Pythagorean Theorem to find r:     x² + y² = r²

19) x = [tex]-\sqrt{17}[/tex] ,  y = 8,    r = 9

     [tex]\cos \theta=\dfrac{x}{r} \rightarrow \quad \cos \theta =\dfrac{-\sqrt{17}}{9} \rightarrow \quad \theta = cos^{-1}\bigg(\dfrac{-\sqrt{17}}{9}\bigg)\rightarrow \quad \theta = \large\boxed{117^o}[/tex]

20) x = 3,   y = 4,   r = 5

      [tex]\cos \theta=\dfrac{x}{r} \rightarrow \quad \cos \theta =\dfrac{3}{5} \rightarrow \quad \theta = cos^{-1}\bigg(\dfrac{3}{5}\bigg)\rightarrow \quad \theta = \large\boxed{53^o}[/tex]

21) x = [tex]-\sqrt7[/tex],   y = -3,   r = 4

    [tex]\sin \theta=\dfrac{y}{r} \rightarrow \quad \sin \theta =\dfrac{-3}{4} \rightarrow \quad \theta = sin^{-1}\bigg(\dfrac{-3}{4}\bigg)\rightarrow \quad \theta = \large\boxed{229^o}[/tex]    

22) x = [tex]-\sqrt{15}[/tex],   y = 7,   r = 8

     [tex]\tan \theta=\dfrac{y}{x} \rightarrow \quad \tan \theta =\dfrac{7}{-\sqrt{15}} \rightarrow \quad \theta = tan^{-1}\bigg(\dfrac{7}{-\sqrt{15}}\bigg)\rightarrow \quad \theta = \large\boxed{119^o}[/tex]

23) x = -13,   y = 6,   r = [tex]\sqrt{205}[/tex]

     [tex]\tan \theta=\dfrac{y}{x} \rightarrow \quad \tan \theta =\dfrac{6}{-13} \rightarrow \quad \theta = tan^{-1}\bigg(\dfrac{6}{-13}\bigg)\rightarrow \quad \theta = \large\boxed{155^o}[/tex]

24) x = 4,    y = -3,   r = 5

     [tex]\sin \theta=\dfrac{y}{r} \rightarrow \quad \sin \theta =\dfrac{-3}{5} \rightarrow \quad \theta = sin^{-1}\bigg(\dfrac{-3}{5}\bigg)\rightarrow \quad \theta = \large\boxed{323^o}[/tex]

batolisis

For the point [tex](-\sqrt{17},8)[/tex], [tex]cos\theta=0.4581[/tex]

For the point [tex](3,4)[/tex], [tex]cos\theta=0.6[/tex]

For the point [tex](-\sqrt7,-3)[/tex], [tex]sin\theta=-0.75[/tex]

For the point [tex](-\sqrt{15},7)[/tex], [tex]tan\theta=-1.807[/tex]

For the point [tex](-13,6)[/tex], [tex]tan\theta=-0.4615[/tex]

For the point [tex](4,-3)[/tex], [tex]sin\theta=-0.6[/tex]

For ratios [tex]sin\theta[/tex] and [tex]cos\theta[/tex], we need to calculate the hypotenuse of the right-angled triangle. For the ratio [tex]tan\theta[/tex], we can directly make use of the coordinate points.

For the point [tex](-\sqrt{17},8)[/tex]

[tex]r=\sqrt{17+8^2}=9[/tex]

[tex]cos\theta=\dfrac{x}{r}\\\\=-\dfrac{\sqrt{17}}{9}\approx0.4581[/tex]

For the point [tex](3,4)[/tex]

[tex]r=\sqrt{3^2+4^2}=5[/tex]

[tex]cos\theta=\dfrac{x}{r}\\\\=\dfrac{3}{5}=0.6[/tex]

For the point [tex](-\sqrt7,-3)[/tex]

[tex]r=\sqrt{7+(-3)^2}=4[/tex]

[tex]sin\theta=\dfrac{y}{r}\\=-\dfrac{3}{4}=-0.75[/tex]

For the point [tex](-\sqrt{15},7)[/tex]

[tex]tan\theta=\dfrac{y}{x}\\=-\dfrac{7}{\sqrt{15}}=-1.807[/tex]

For the point [tex](-13,6)[/tex]

[tex]tan\theta=\dfrac{y}{x}\\=-\dfrac{6}{13}=-0.4615[/tex]

For the point [tex](4,-3)[/tex]

[tex]r=\sqrt{4^2+(-3)^2}=5[/tex]

[tex]sin\theta=\dfrac{y}{r}\\=-\dfrac{3}{5}=-0.6[/tex]

Learn more about trigonometry here: https://brainly.com/question/23686339

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