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)=(13)/(12)`

Recall the trigonometric identity:

`\csc (\theta)=(1)/(\sin (\theta))`

Therefore, we have that

`\sin (\theta)=(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)=(5)/(13)`