Question

find the cosine of ∠y. diagram: right triangle wyx with right angle at w, wx = 10, yx = 10√2 write your answer in simplified, rationalized form. do not round. cos(y) =

Explanation:

Step1: Identify triangle type and sides

Triangle \( WYX \) is right - angled at \( W \). So, it is a right - triangle. In a right - triangle, for an acute angle \( \theta \), \( \cos\theta=\frac{\text{adjacent side}}{\text{hypotenuse}} \). For \( \angle Y \), we need to find the adjacent side and the hypotenuse.

First, we can find the length of \( WY \) using the Pythagorean theorem. The Pythagorean theorem states that in a right - triangle \( a^{2}+b^{2}=c^{2} \), where \( c \) is the hypotenuse and \( a,b \) are the legs. Here, hypotenuse \( XY = 10\sqrt{2} \), one leg \( WX = 10 \), and let the other leg \( WY=x \).

So, \( WY^{2}+WX^{2}=XY^{2} \)
\( x^{2}+10^{2}=(10\sqrt{2})^{2} \)
\( x^{2}+100 = 100\times2 \)
\( x^{2}+100 = 200 \)
\( x^{2}=200 - 100=100 \)
\( x = 10 \) (since length cannot be negative)

Step2: Calculate \( \cos(Y) \)

For \( \angle Y \), the adjacent side (the side adjacent to \( \angle Y \)) is \( WY = 10 \) and the hypotenuse is \( XY=10\sqrt{2} \).

So, \( \cos(Y)=\frac{\text{adjacent to }Y}{\text{hypotenuse}}=\frac{WY}{XY}=\frac{10}{10\sqrt{2}} \)

Simplify the fraction: \( \frac{10}{10\sqrt{2}}=\frac{1}{\sqrt{2}} \)

Rationalize the denominator: Multiply the numerator and denominator by \( \sqrt{2} \), we get \( \frac{1\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}}=\frac{\sqrt{2}}{2} \)

Answer:

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