Question

use the figure below to evaluate each trigonometric function of angle a. answer exactly.
question help: video message instructor

Explanation:

Step1: Find the hypotenuse

By the Pythagorean theorem \(c=\sqrt{a^{2}+b^{2}}\), where \(a = 3\) and \(b=9\).

$$c=\sqrt{3^{2}+9^{2}}=\sqrt{9 + 81}=\sqrt{90}=3\sqrt{10}$$

Step2: Calculate \(\sin(A)\)

\(\sin(A)=\frac{\text{opposite}}{\text{hypotenuse}}\). The side opposite to angle \(A\) is \(9\).
\(\sin(A)=\frac{9}{3\sqrt{10}}=\frac{3}{\sqrt{10}}=\frac{3\sqrt{10}}{10}\)

Step3: Calculate \(\cos(A)\)

\(\cos(A)=\frac{\text{adjacent}}{\text{hypotenuse}}\). The side adjacent to angle \(A\) is \(3\).
\(\cos(A)=\frac{3}{3\sqrt{10}}=\frac{1}{\sqrt{10}}=\frac{\sqrt{10}}{10}\)

Step4: Calculate \(\tan(A)\)

\(\tan(A)=\frac{\text{opposite}}{\text{adjacent}}\)
\(\tan(A)=\frac{9}{3}=3\)

Step5: Calculate \(\cot(A)\)

\(\cot(A)=\frac{\text{adjacent}}{\text{opposite}}\)
\(\cot(A)=\frac{3}{9}=\frac{1}{3}\)

Step6: Calculate \(\sec(A)\)

\(\sec(A)=\frac{\text{hypotenuse}}{\text{adjacent}}\)
\(\sec(A)=\frac{3\sqrt{10}}{3}=\sqrt{10}\)

Step7: Calculate \(\csc(A)\)

\(\csc(A)=\frac{\text{hypotenuse}}{\text{opposite}}\)
\(\csc(A)=\frac{3\sqrt{10}}{9}=\frac{\sqrt{10}}{3}\)

Answer:

\(\sin(A)=\frac{3\sqrt{10}}{10}\), \(\cos(A)=\frac{\sqrt{10}}{10}\), \(\tan(A) = 3\), \(\cot(A)=\frac{1}{3}\), \(\sec(A)=\sqrt{10}\), \(\csc(A)=\frac{\sqrt{10}}{3}\)