Q:

# The point (-7, -24) is on the terminal ray of angle θ which is in standard position. A student found the six trigonometric values for angle θ. The student’s answers are shown

Accepted Solution

A:
$$\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad cos(\theta)=\cfrac{adjacent}{hypotenuse} \quad % tangent tan(\theta)=\cfrac{opposite}{adjacent} \\\\\\ % cotangent cot(\theta)=\cfrac{adjacent}{opposite} \qquad % cosecant csc(\theta)=\cfrac{hypotenuse}{opposite} \quad % secant sec(\theta)=\cfrac{hypotenuse}{adjacent}\\\\ -------------------------------$$

$$\bf (\stackrel{a}{-7}~~,~~\stackrel{b}{-24}) \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{(-7)^2+(-24)^2}\implies c=\sqrt{49+576}\implies c=\sqrt{625} \\\\\\ c=25\impliedby hypotenuse\\\\ -------------------------------$$

$$\bf sin(\theta )=\cfrac{-24}{25}\qquad cos(\theta )=\cfrac{-7}{25}\qquad \qquad tan(\theta )=\cfrac{-24}{-7}\implies \cfrac{24}{7} \\\\\\ cot(\theta )=\cfrac{-7}{-24}\implies \cfrac{7}{24}\qquad \qquad sec(\theta )=\cfrac{25}{-7}\qquad csc(\theta )=\cfrac{25}{-24}$$