Math / Trigonometry

Question10. Find the exact value of $\cos I$ in simplest radical form.

Studdy Solution

STEP 1

1. The triangle $\triangle IJK$ is a right triangle with a right angle at $J$ .
2. The side lengths are given as $IJ = 9$ , $JK = \sqrt{8}$ , and $IK = \sqrt{89}$ .
3. We need to find the cosine of angle $I$ .

STEP 2

1. Verify the triangle using the Pythagorean theorem.
2. Use the definition of cosine for angle $I$ .

STEP 3

Verify the triangle using the Pythagorean theorem. The Pythagorean theorem states that for a right triangle:
$a^2 + b^2 = c^2$
where $c$ is the hypotenuse. Here, $IJ = 9$ , $JK = \sqrt{8}$ , and $IK = \sqrt{89}$ is the hypotenuse. Check:
$9^2 + (\sqrt{8})^2 = (\sqrt{89})^2$
Calculate:
$81 + 8 = 89$
$89 = 89$
The triangle satisfies the Pythagorean theorem.

STEP 4

Use the definition of cosine for angle $I$ . In a right triangle, the cosine of an angle is defined as the ratio of the length of the adjacent side to the hypotenuse. For angle $I$ :
$\cos I = \frac{\text{Adjacent side to } I}{\text{Hypotenuse}}$
Here, the adjacent side to angle $I$ is $JK = \sqrt{8}$ , and the hypotenuse is $IK = \sqrt{89}$ .

STEP 5

Substitute the values into the cosine formula:
$\cos I = \frac{\sqrt{8}}{\sqrt{89}}$
Simplify the expression if possible. Since both are square roots, this is already in simplest radical form.
The exact value of $\cos I$ in simplest radical form is:
$\boxed{\frac{\sqrt{8}}{\sqrt{89}}}$

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