Question

If cos=2/3 find csc; if 0°<0<90°

Answer 1

\(\bf cos(\theta)=\cfrac{2}{3}\qquad 0^o<\theta<90^o
\\ \quad \\
0^o<\theta<90^o\Leftarrow\textit{another way of saying, "a" and "b" are positive}
\\ \quad \\
cos(\theta)=\cfrac{2}{3}\implies \cfrac{adjacent}{hypotenuse}\implies \cfrac{a=2}{c=3}\qquad \textit{so what's "b"?}
\\ \quad \\
c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}={\color{blue}{ b}}\qquad csc(\theta)=\cfrac{c}{{\color{blue}{ b}}}\)

the pythagorean theorem doesn't tell us if the root value is +/-, so what's "b"?
well we already know is in the I Quadrant, so is positive