bearing in mind that the hypotenuse is never negative, since it's just a distance unit from the center of the circle, we know the sine is -(4/5), since the hypotenuse is positive, then the negative must be the 4, or -4/5.
we know the angle θ is in the IV Quadrant, where the x/adjacent is positive and the y/opposite is negative, so

![\bf \pm√(5^2-(-4)^2)=a\implies \pm√(25-16)=a\implies \pm 3=a\implies \stackrel{IV~Quadrant}{+3=a} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill sec(\theta )= \cfrac{\stackrel{hypotenuse}{5}}{\stackrel{adjacent}{3}}~\hfill](https://img.qammunity.org/2020/formulas/mathematics/middle-school/dq6jfs3oxape2p94h3kkkjumfmkk4pi4uw.png)