Question

find the cosine of $\angle h$.

$f$ $8\sqrt{2}$ $h$

$8$

$g$

write your answer in simplified, rationalized form. do not round.

$\cos (h) = \square$

Explanation:

Step1: Identify triangle type

Triangle \( FGH \) is right - angled at \( G \), so it is a right - triangle. In a right - triangle, we can use the definitions of trigonometric ratios. The cosine of an acute angle \( \theta \) in a right - triangle is defined as \( \cos\theta=\frac{\text{adjacent side to } \theta}{\text{hypotenuse}} \).

Step2: Determine sides for \( \angle H \)

First, we need to find the length of side \( GH \). Since \( \triangle FGH \) is a right - triangle with \( \angle G = 90^{\circ} \), \( FG = 8 \), and \( FH=8\sqrt{2} \), we can use the Pythagorean theorem \( a^{2}+b^{2}=c^{2} \) (where \( c \) is the hypotenuse and \( a,b \) are the legs). Let \( FG = a = 8 \), \( GH = b \), and \( FH=c = 8\sqrt{2} \). Then \( b=\sqrt{c^{2}-a^{2}} \).

Substitute the values: \( b=\sqrt{(8\sqrt{2})^{2}-8^{2}}=\sqrt{128 - 64}=\sqrt{64} = 8 \).

Now, for \( \angle H \), the adjacent side to \( \angle H \) is \( GH = 8 \) and the hypotenuse is \( FH=8\sqrt{2} \).

Step3: Calculate \( \cos(H) \)

Using the definition of cosine, \( \cos(H)=\frac{\text{adjacent to }H}{\text{hypotenuse}}=\frac{GH}{FH} \).

Substitute \( GH = 8 \) and \( FH = 8\sqrt{2} \): \( \cos(H)=\frac{8}{8\sqrt{2}} \).

Simplify the fraction: Cancel out the common factor of 8 in the numerator and the denominator, we get \( \frac{1}{\sqrt{2}} \). Rationalize the denominator by multiplying the numerator and the denominator by \( \sqrt{2} \): \( \frac{1\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}}=\frac{\sqrt{2}}{2} \).

Answer:

\(\frac{\sqrt{2}}{2}\)