Question

use the figure below to evaluate each trigonometric function of angle a. answer exactly.

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Explanation:

Step1: Find the hypotenuse

Using the Pythagorean theorem \(c=\sqrt{a^{2}+b^{2}}\), where \(a = 6\) and \(b=12\).
\(c=\sqrt{6^{2}+12^{2}}=\sqrt{36 + 144}=\sqrt{180}=6\sqrt{5}\)

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

\(\sin(A)=\frac{\text{opposite}}{\text{hypotenuse}}\), the side opposite to angle \(A\) is \(12\)
\(\sin(A)=\frac{12}{6\sqrt{5}}=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}\)

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

\(\cos(A)=\frac{\text{adjacent}}{\text{hypotenuse}}\), the side adjacent to angle \(A\) is \(6\)
\(\cos(A)=\frac{6}{6\sqrt{5}}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}\)

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

\(\tan(A)=\frac{\text{opposite}}{\text{adjacent}}\)
\(\tan(A)=\frac{12}{6} = 2\)

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

\(\cot(A)=\frac{\text{adjacent}}{\text{opposite}}\)
\(\cot(A)=\frac{6}{12}=\frac{1}{2}\)

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

\(\sec(A)=\frac{1}{\cos(A)}\)
\(\sec(A)=\sqrt{5}\)

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

\(\csc(A)=\frac{1}{\sin(A)}\)
\(\csc(A)=\frac{\sqrt{5}}{2}\)

Answer:

\(\sin(A)=\frac{2\sqrt{5}}{5}\), \(\cos(A)=\frac{\sqrt{5}}{5}\), \(\tan(A)=2\), \(\cot(A)=\frac{1}{2}\), \(\sec(A)=\sqrt{5}\), \(\csc(A)=\frac{\sqrt{5}}{2}\)