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find the cosine of ∠v. triangle with right angle at u, vu=12, ut=35 sim…

Question

find the cosine of ∠v.

triangle with right angle at u, vu=12, ut=35

simplify your answer and write it as a proper fraction, improper fraction, or whole number.

cos(v) = blank box

buttons for fraction and square root

Explanation:

Step1: Identify the triangle type and sides

This is a right - triangle \( \triangle VUT \) with \( \angle U = 90^{\circ} \). For \( \angle V \), we need to find the adjacent side, opposite side and hypotenuse. The adjacent side to \( \angle V \) is \( VU = 12 \), the opposite side is \( UT=35 \). First, we find the hypotenuse \( VT \) using the Pythagorean theorem \( a^{2}+b^{2}=c^{2} \), where \( a = 12 \), \( b = 35 \) and \( c=VT \).
\( VT=\sqrt{12^{2}+35^{2}}=\sqrt{144 + 1225}=\sqrt{1369}=37 \)

Step2: Recall the definition of cosine

The cosine of an angle in a right - triangle is defined as the ratio of the adjacent side to the hypotenuse. For \( \angle V \), \( \cos(V)=\frac{\text{Adjacent side to } \angle V}{\text{Hypotenuse}} \)
The adjacent side to \( \angle V \) is \( VU = 12 \) and the hypotenuse \( VT = 37 \). So \( \cos(V)=\frac{12}{37} \)

Answer:

\(\frac{12}{37}\)