MATH SOLVE

7 months ago

Q:
# Given that tan 0=-4/7 , and 270 degrees <0<360 degrees , what is the exact value of sec0?

Accepted Solution

A:

I take it you meant θ angle, anyway.

we know the tan(θ) = -4/7... alrite, we also know that 270° < θ < 360°, which is another to say that θ is in the IV quadrant, where the adjacent side or "x" value is positive whilst the opposite side or "y" value is negative.

[tex]\bf tan(\theta )=\cfrac{\stackrel{opposite}{-4}}{\stackrel{adjacent}{7}}\impliedby \textit{now let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{7^2+(-4)^2}\implies \implies c=\sqrt{65}\\\\ -------------------------------\\\\ sec(\theta )=\cfrac{\stackrel{hypotenuse}{\sqrt{65}}}{\stackrel{adjacent}{7}}[/tex]

we know the tan(θ) = -4/7... alrite, we also know that 270° < θ < 360°, which is another to say that θ is in the IV quadrant, where the adjacent side or "x" value is positive whilst the opposite side or "y" value is negative.

[tex]\bf tan(\theta )=\cfrac{\stackrel{opposite}{-4}}{\stackrel{adjacent}{7}}\impliedby \textit{now let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{7^2+(-4)^2}\implies \implies c=\sqrt{65}\\\\ -------------------------------\\\\ sec(\theta )=\cfrac{\stackrel{hypotenuse}{\sqrt{65}}}{\stackrel{adjacent}{7}}[/tex]