Respuesta :
If the reference angle is theta and, we are given;
[tex]\sin \theta=-\frac{3}{5}[/tex]Note that sin theta is negative.
Also, we have the following on the unit circle;
[tex]\begin{gathered} \cos \theta=\frac{x}{r} \\ \sin \theta=\frac{y}{r} \end{gathered}[/tex]Where the value of y is negative, we would have a negative sin theta, that is;
[tex]\sin \theta=-\frac{y}{r}[/tex]This only occurs where y is negative, which is in the third and fourth quadrants. Therefore,
(a) The reference angle theta could be either in the third or fourth quadrants.
Note also that
[tex]\sec \theta=\frac{1}{\cos \theta}[/tex]To determine the value of the cosine of this angle, we shall take the given values, y and r.
Note that r is the hypotenuse, while y is the height. To find the base x,
[tex]\begin{gathered} r^2=x^2+y^2 \\ 5^2=x^2+3^2 \\ 25=x^2+9 \\ \text{Subtract 9 from both sides;} \\ 16=x^2 \\ \text{Take the square root of both sides;} \\ 4=x \end{gathered}[/tex]Where the cosine is given as;
[tex]\begin{gathered} \cos \theta=\frac{x}{r} \\ \cos \theta=\frac{4}{5} \\ \text{Therefore,} \\ \sec \theta=\frac{1}{\cos \theta} \\ \sec \theta=\frac{1}{\frac{4}{5}} \\ \sec \theta=\frac{5}{4} \end{gathered}[/tex]