The vector is given to be:
[tex]t=-8i+6j[/tex]For a vector v given to be:
[tex]v=ai+bj[/tex]the trigonometric form is given to be:
[tex]v=|v|\langle\cos \theta,\sin \theta\rangle[/tex]where
[tex]|v|=\sqrt[]{a^2+b^2}[/tex]and
[tex]\theta=\tan ^{-1}\frac{b}{a}[/tex]Comparing this to our vector, we have:
[tex]\begin{gathered} a=-8 \\ b=6 \end{gathered}[/tex]Therefore,
[tex]\begin{gathered} |t|=\sqrt[]{(-8)^2+6^2}=\sqrt[]{64+36}=\sqrt[]{100} \\ |t|=10 \end{gathered}[/tex]and
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{6}{-8})=_{}\tan ^{-1}(-\frac{6}{8})_{} \\ \theta=-36.87 \end{gathered}[/tex]Since the angle is negative, we can add 180° to the angle to get the positive angle. Therefore:
[tex]\theta=-36.87+180=143.13[/tex]Therefore, the vector in trigonometric form will be:
[tex]t=10\langle\cos 143.13,\sin 143.13\rangle[/tex]The correct option is the SECOND OPTION.