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.

Respuesta :

Given in the question is:

[tex]\csc (\theta)=\frac{13}{12}[/tex]

Recall the trigonometric identity:

[tex]\csc (\theta)=\frac{1}{\sin (\theta)}[/tex]

Therefore, we have that

[tex]\sin (\theta)=\frac{12}{13}[/tex]

Recall the trigonometric ratio:

[tex]\begin{gathered} \sin (\theta)=\frac{\text{opp}}{\text{hyp}} \\ \cos (\theta)=\frac{\text{adj}}{\text{hyp}} \end{gathered}[/tex]

and, using the Pythagorean Theorem:

[tex]hyp^2=opp^2+adj^2[/tex]

From the sin value, we have:

[tex]\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}[/tex]

Therefore, the value of cos(θ) is:

[tex]\sin (\theta)=\frac{5}{13}[/tex]