Question

find the cosine of \\( \angle v \\).
simplify your answer and write it as a proper fraction, improper fraction, or whole number.
\\( \cos ( v ) = \\)

Explanation:

Step1: Recall the cosine formula

In a right - triangle, \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). For \(\angle V\), the adjacent side to \(\angle V\) is \(VW\), and the hypotenuse is \(VU\).

Step2: Identify the lengths

We are given that \(VU = 89\) (hypotenuse) and \(VW\) needs to be found. Using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\) (where \(c = 89\), \(b = 80\), and \(a=VW\)), we have \(VW=\sqrt{89^{2}-80^{2}}=\sqrt{(89 + 80)(89 - 80)}=\sqrt{169\times9}=\sqrt{1521} = 39\). But wait, no! Wait, the formula \(\cos V=\frac{VW}{VU}\). Wait, no, wait the sides: in right - triangle \(VUW\) with right - angle at \(W\), for \(\angle V\), adjacent side \(VW\), hypotenuse \(VU\). Given \(VU = 89\), \(VW\): no, wait, no! Wait, the formula \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). For \(\angle V\), adjacent is \(VW\), hypotenuse is \(VU\). Wait, no: \(\cos V=\frac{VW}{VU}\). Wait, \(VU = 89\), \(VW\): using Pythagorean theorem \(VW=\sqrt{VU^{2}-UW^{2}}\). Wait no! Wait, \(\cos V=\frac{VW}{VU}\). Given \(VU = 89\), \(VW\): Wait, no, the sides: \(VU = 89\) (hypotenuse), \(UW = 80\) (opposite to \(\angle V\)), \(VW\) (adjacent to \(\angle V\)). By Pythagorean theorem \(VW=\sqrt{89^{2}-80^{2}}=\sqrt{(89 + 80)(89 - 80)}=\sqrt{169\times9}=\sqrt{1521}=39\). Then \(\cos V=\frac{VW}{VU}=\frac{39}{89}\).

Answer:

\(\frac{39}{89}\)