Question

find the cosine of ∠i.
triangle with vertices j, h, i; right angle at j; jh = √41; ji = √14
write your answer in simplified, rationalized form. do not round.
cos(i) = blank box fraction and square root buttons

Explanation:

Step1: Recall cosine definition

In a right triangle, $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$. For $\angle I$, adjacent side is $IJ = \sqrt{14}$, hypotenuse is $IH$. First, find hypotenuse $IH$ using Pythagorean theorem: $IH^2 = IJ^2 + JH^2$.
$$IH^2 = (\sqrt{14})^2 + (\sqrt{41})^2 = 14 + 41 = 55$$
So $IH = \sqrt{55}$.

Step2: Calculate $\cos(I)$

Adjacent to $\angle I$ is $IJ = \sqrt{14}$, hypotenuse is $IH = \sqrt{55}$. Thus:
$$\cos(I) = \frac{\sqrt{14}}{\sqrt{55}}$$
Rationalize the denominator: multiply numerator and denominator by $\sqrt{55}$:
$$\cos(I) = \frac{\sqrt{14} \times \sqrt{55}}{\sqrt{55} \times \sqrt{55}} = \frac{\sqrt{770}}{55}$$

Answer:

$\frac{\sqrt{770}}{55}$