Question

find sec θ.
(-5, √11)
r
θ
sec θ = -?/

Explanation:

Step1: Recall the formula for \( r \)

For a point \((x, y)\) in the coordinate plane, \( r=\sqrt{x^{2}+y^{2}} \). Here, \( x = - 5 \) and \( y=\sqrt{11} \). So, \( r=\sqrt{(-5)^{2}+(\sqrt{11})^{2}}=\sqrt{25 + 11}=\sqrt{36}=6 \).

Step2: Recall the definition of \( \sec\theta \)

\( \sec\theta=\frac{r}{x} \) (since \( \cos\theta=\frac{x}{r} \) and \( \sec\theta=\frac{1}{\cos\theta} \)). Substituting \( r = 6 \) and \( x=-5 \), we get \( \sec\theta=\frac{6}{-5}=-\frac{6}{5} \).

Answer:

\(\frac{6}{5}\) (the numerator is 6, the denominator is 5)