Given the figure of a right-angle triangle and the indicated angle θ
As shown, the hypotenuse = 7
And the opposite side = 4
We will use the Pythagorean theroem to find the adjacent side:
[tex]adjacent=\sqrt{7^2-4^2}=\sqrt{49-16}=\sqrt{33}[/tex]
Now, we will find cot θ, sec θ, and cos θ as follows:
[tex]\begin{gathered} \cotθ=\frac{adjacent}{opposite}=\frac{\sqrt{33}}{4} \\ \\ \secθ=\frac{hypotenuse}{adjacent}=\frac{7}{\sqrt{33}}=\frac{7\sqrt{33}}{33} \\ \\ \cosθ=\frac{adjacent}{hypotenuse}=\frac{\sqrt{33}}{7} \end{gathered}[/tex]
So, the answer will be:
[tex]\begin{gathered} \cotθ=\frac{\sqrt{33}}{4} \\ \\ \secθ=\frac{7\sqrt{33}}{33} \\ \\ \cosθ=\frac{\sqrt{33}}{7} \end{gathered}[/tex]
