Question

find the secant of $\angle i$.
write your answer in simplified, rationalized form. do not round.
$\sec(i)=$

Explanation:

Step1: Find the adjacent side to \(\angle I\)

Use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\). Let \(JH = 2\sqrt{2}\) (one - leg), \(IJ=\sqrt{17}\) (hypotenuse), and \(IH\) (the other leg). Then \(IH=\sqrt{IJ^{2}-JH^{2}}=\sqrt{17-(2\sqrt{2})^{2}}=\sqrt{17 - 8}=\sqrt{9}=3\). The adjacent side to \(\angle I\) is \(IH = 3\), and the hypotenuse is \(IJ=\sqrt{17}\).

Step2: Recall the definition of secant

The secant of an angle in a right - triangle is defined as \(\sec\theta=\frac{\text{hypotenuse}}{\text{adjacent}}\). For \(\angle I\), \(\sec(I)=\frac{IJ}{IH}\).

Step3: Substitute the values

Substitute \(IJ = \sqrt{17}\) and \(IH = 3\) into the formula: \(\sec(I)=\frac{\sqrt{17}}{3}\).

Answer:

\(\frac{\sqrt{17}}{3}\)