Question

find: \\(\cos f\\) (and there is a right triangle (efg) with right angle at (e), (eg = 35), (fg = 37))

Explanation:

Step1: Identify the right triangle

We have a right triangle \( \triangle EFG \) with \( \angle E = 90^\circ \), \( EG = 35 \), and \( FG = 37 \). First, we need to find the length of \( EF \) 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, \( FG \) is the hypotenuse, so \( EF^2 + EG^2 = FG^2 \).
\[
EF^2 + 35^2 = 37^2
\]
\[
EF^2 = 37^2 - 35^2
\]
We can use the difference of squares formula \( a^2 - b^2=(a + b)(a - b) \), where \( a = 37 \) and \( b = 35 \).
\[
EF^2=(37 + 35)(37 - 35)=(72)(2)=144
\]
Taking the square root of both sides, \( EF=\sqrt{144} = 12 \).

Step2: Recall the definition of cosine in a right triangle

In a right triangle, the cosine of an acute angle is the ratio of the adjacent side to the hypotenuse. For \( \angle F \), the adjacent side is \( EF \) and the hypotenuse is \( FG \). So, \( \cos F=\frac{\text{Adjacent}}{\text{Hypotenuse}}=\frac{EF}{FG} \).

Step3: Substitute the values of \( EF \) and \( FG \)

We know that \( EF = 12 \) and \( FG = 37 \). So,
\[
\cos F=\frac{12}{37}
\]

Answer:

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