Respuesta :

SOLUTION

We are told that

[tex]\begin{gathered} \sin \theta=\frac{3}{5} \\ i\text{n the second quadrant } \end{gathered}[/tex]

In the second quadrant the required angle =

[tex]180-\theta[/tex]

Also only sin is positive in the second quadrant

So

[tex]\begin{gathered} \text{sin}\theta\text{ = sin(180 -}\theta) \\ \text{sin(180 -}\theta)=\frac{3}{5} \\ \text{(180 -}\theta)=\sin ^{-1}\frac{3}{5} \\ 180-\theta=36.86989765 \\ \theta=180-36.86989765 \\ \theta=143.1301024 \end{gathered}[/tex]

(a)

[tex]\begin{gathered} \cos \theta=-cos(180-\theta) \\ -cos(180-\theta)=-\cos (180-143.1301024) \\ =-cos36.86989765 \\ =-\frac{4}{5} \end{gathered}[/tex]

(b)

[tex]\begin{gathered} \tan \theta=-\tan 36.86989765 \\ =-\frac{3}{4} \end{gathered}[/tex]

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