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Sagot :
Given the point (-9,-4) that is on on the terminal side of an angle in standard position.
[tex]x=-9,y=-4[/tex]First, we determine the value of r, the hypotenuse.
[tex]\begin{gathered} r^2=(-9)^2+(-4)^2 \\ r^2=81+16 \\ r^2=97 \\ r=\sqrt{97} \end{gathered}[/tex](a) Sin (θ)
[tex]\begin{gathered} \sin \theta=\frac{Opposite}{\text{Hypotenuse}} \\ =\frac{y}{r} \\ =\frac{-4}{\sqrt{97}} \\ =-\frac{4\sqrt{97}}{97} \end{gathered}[/tex](b) Cos (θ)
[tex]\begin{gathered} \cos \theta=\frac{\text{Adjacent}}{\text{Hypotenuse}} \\ =\frac{x}{r} \\ =\frac{-9}{\sqrt{97}} \\ =-\frac{9\sqrt{97}}{97} \end{gathered}[/tex](c) Tan (θ)
[tex]\begin{gathered} \tan \theta=\frac{\text{Opposite}}{\text{Adjacent}} \\ =\frac{y}{x} \\ =\frac{-4}{-9} \\ =\frac{4}{9} \end{gathered}[/tex](d) csc (θ)
[tex]\begin{gathered} \cosec \theta=\frac{1}{\sin \theta} \\ =-\frac{\sqrt{97}}{4} \end{gathered}[/tex](e)sec(θ)
[tex]\begin{gathered} \sec \theta=\frac{1}{\cos \theta} \\ =-\frac{\sqrt{97}}{9} \end{gathered}[/tex]
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