Question

Help me please so i can see if i’m on the rights track. if csc (θ) = 13/12 and 0° < θ < 90°, what is cos (θ)? write the answer in simplified, rationalized form.

Answer 1

Given in the question is:

$\csc (\theta)=\frac{13}{12}$

Recall the trigonometric identity:

$\csc (\theta)=\frac{1}{\sin (\theta)}$

Therefore, we have that

$\sin (\theta)=\frac{12}{13}$

Recall the trigonometric ratio:

$\begin{gathered} \sin (\theta)=\frac{\text{opp}}{\text{hyp}} \\ \cos (\theta)=\frac{\text{adj}}{\text{hyp}} \end{gathered}$

and, using the Pythagorean Theorem:

$hyp^2=opp^2+adj^2$

From the sin value, we have:

$\begin{gathered} opp=12 \\ hyp=13 \\ \therefore \\ 13^2=12^2+adj^2 \\ 169=144+adj^2 \\ adj^2=169-144=25 \\ adj=\sqrt[]{25} \\ adj=5 \end{gathered}$

Therefore, the value of cos(θ) is:

$\sin (\theta)=\frac{5}{13}$