Question

Given that θ terminates in Quadrant II and sinθ = 12/13 , find secθ.

Answer 1

Keeping in mind that, on the II Quadrant, the sine/opposite is positive whilst the cosine/adjacent is negative, then

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{12}}{\stackrel{hypotenuse}{13}}\impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-12^2}=a\implies \pm\sqrt{25}=a \\\\\\ \pm 5=a\implies \stackrel{II~Quadrant}{-5=a}\qquad therefore\qquad sec(\theta )=\cfrac{\stackrel{hypotenuse}{13}}{\stackrel{adjacent}{-5}}$