Question

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

Explanation:

Step1: Recall cosine definition

In a right triangle, $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$ for angle $\theta$.
For $\angle G$, adjacent side is $GF = 24$? Wait, no, wait. Wait, right triangle $FGH$ with right angle at $F$. So sides: $FG = 24$, $FH = 70$, hypotenuse $GH = 74$. Wait, angle at $G$: adjacent side to $\angle G$ is $FG$? No, wait, in triangle $FGH$, right-angled at $F$. So vertices: $F$ (right angle), $G$, $H$. So sides: $FG$ (one leg), $FH$ (another leg), $GH$ (hypotenuse). So for $\angle G$: the sides: adjacent is $FG$? Wait, no. Let's label: right angle at $F$, so $F$ is between $G$ and $H$? Wait, the diagram: $F$ has right angle, $G$ and $H$ are the other two vertices. So $FG$ is a leg, $FH$ is a leg, $GH$ is hypotenuse. So for angle $G$: the sides: opposite side is $FH = 70$, adjacent side is $FG = 24$, hypotenuse is $GH = 74$. Wait, no, wait: in angle $G$, the sides: adjacent is the leg next to $G$ (not the hypotenuse), so $FG$ is adjacent? Wait, no, let's think again. In triangle $FGH$, right-angled at $F$. So angle at $G$: the sides: adjacent side is $FG$ (since $FG$ is one leg, and $G$ is at vertex $G$, so the sides forming angle $G$ are $FG$ and $GH$. Wait, no: angle at $G$ is between $FG$ and $GH$? Wait, no, the triangle is $F - G - H$? No, the right angle is at $F$, so $F$ is connected to $G$ and $H$, and $G$ and $H$ are connected. So the sides: $FG$ (from $F$ to $G$), $FH$ (from $F$ to $H$), $GH$ (from $G$ to $H$). So angle at $G$: the two sides are $FG$ (length 24) and $GH$ (hypotenuse, length 74), and the opposite side is $FH$ (length 70). Wait, no, adjacent side to angle $G$ is $FG$? Wait, no, adjacent side is the side that is part of the angle and is not the hypotenuse. So for angle $G$, the adjacent side is $FG$ (length 24), and the hypotenuse is $GH$ (length 74). Wait, but wait, let's check Pythagoras: $24^2 + 70^2 = 576 + 4900 = 5476$, and $74^2 = 5476$. So that's correct. So $\cos(G) = \frac{\text{adjacent to } G}{\text{hypotenuse}} = \frac{FG}{GH} = \frac{24}{74}$? Wait, no, wait, that can't be. Wait, no, I think I mixed up the adjacent side. Wait, angle at $G$: the sides: the leg adjacent to $G$ is $FG$? Wait, no, maybe I got the adjacent and opposite wrong. Wait, let's use the definition: in a right triangle, for angle $\theta$, $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$, where adjacent is the side that is adjacent (next to) the angle, and hypotenuse is the longest side (opposite the right angle). So in triangle $FGH$, right-angled at $F$, so hypotenuse is $GH = 74$. Angle at $G$: the sides: adjacent side is $FG$ (length 24), opposite side is $FH$ (length 70). Wait, but then $\cos(G) = \frac{FG}{GH} = \frac{24}{74}$? But that simplifies to $\frac{12}{37}$. Wait, but wait, maybe I made a mistake. Wait, no, wait: maybe the adjacent side is $FH$? No, no. Wait, let's draw the triangle: right angle at $F$, so $F$ is the right angle, so $FG$ and $FH$ are the legs, $GH$ is hypotenuse. So angle at $G$: the sides: $FG$ is one leg (from $F$ to $G$), $GH$ is hypotenuse (from $G$ to $H$), and $FH$ is the other leg (from $F$ to $H$). So for angle $G$, the adjacent side is $FG$ (because it's the leg that's part of angle $G$ along with the hypotenuse), and the opposite side is $FH$. So $\cos(G) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{FG}{GH} = \frac{24}{74}$. Wait, but $24$ and $74$ have a common factor of 2, so simplify: $\frac{24 \div 2}{74 \div 2} = \frac{12}{37}$. Wait, but wait, that seems small. Wait, maybe I mixed up the adjacent side. Wait, no, wait: maybe the adjacent side is $FH$? No, $FH$ is opposite to angle $G$. Wait, no, angle at $G$: the sides: adjacent is the side next to $G$ that's not the hypotenuse. So $FG$ is next to $G$ (since $G$ is connected to $F$ and $H$), so $FG$ is adjacent, $FH$ is opposite, $GH$ is hypotenuse. So $\cos(G) = \frac{FG}{GH} = \frac{24}{74} = \frac{12}{37}$. Wait, but let's check with Pythagoras: $24^2 + 70^2 = 576 + 4900 = 5476$, and $74^2 = 5476$, so that's correct. So the adjacent side to $G$ is $24$, hypotenuse is $74$, so $\cos(G) = \frac{24}{74} = \frac{12}{37}$.

Step2: Simplify the fraction

$\frac{24}{74}$ can be simplified by dividing numerator and denominator by 2: $\frac{24 \div 2}{74 \div 2} = \frac{12}{37}$.

Answer:

$\frac{12}{37}$