Question

1. what is the cosine ratio for ∠f?

(\frac{13}{5})
(\frac{5}{13})
(\frac{13}{12})
(\frac{12}{13}) (with a right triangle labeled h, g, f, where hg=12, gf=5, hf=13, and right angle at g)

Explanation:

Step1: Recall cosine definition

In a right triangle, $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$ for an angle $\theta$.

Step2: Identify sides for $\angle F$

For $\angle F$ in $\triangle FGH$ (right-angled at $G$):

• Adjacent side to $\angle F$: $FG = 5$
• Hypotenuse: $FH = 13$? Wait, no—wait, let's re-examine. Wait, $\angle F$: the sides. Let's label the triangle: right angle at $G$, so sides: $FG = 5$ (vertical), $HG = 12$ (horizontal), $FH = 13$ (hypotenuse). For $\angle F$, the adjacent side is $FG$? Wait, no. Wait, angle at $F$: the sides. The sides forming $\angle F$ are $FG$ (length 5) and $FH$ (hypotenuse 13), and the opposite side is $HG$ (12). Wait, no—cosine of $\angle F$: adjacent over hypotenuse. The adjacent side to $\angle F$ is the leg that is part of $\angle F$ (other than hypotenuse). So $\angle F$ is at vertex $F$, so the sides: $FG$ (length 5, adjacent to $\angle F$) and $FH$ (hypotenuse 13). Wait, no, wait: in a right triangle, for angle $F$, the adjacent side is the leg connected to $F$ and the right angle? Wait, no. Let's use SOHCAHTOA. SOH: $\sin = \text{opposite}/\text{hypotenuse}$, CAH: $\cos = \text{adjacent}/\text{hypotenuse}$, TOA: $\tan = \text{opposite}/\text{adjacent}$.

For $\angle F$:

• Opposite side: $HG = 12$ (opposite to $\angle F$)
• Adjacent side: $FG = 5$ (adjacent to $\angle F$, since it's the leg forming $\angle F$ with hypotenuse $FH$)
• Hypotenuse: $FH = 13$

So $\cos(\angle F) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{FG}{FH} = \frac{5}{13}$.

Wait, but let's check again. Wait, maybe I mixed up. Let's see: vertex $F$, $G$ is right angle. So sides: $FG = 5$, $HG = 12$, $FH = 13$. So angle at $F$: the sides are $FG$ (length 5) and $FH$ (hypotenuse 13), and the other side is $HG$ (12). So adjacent to $\angle F$ is $FG$ (since it's one of the legs forming $\angle F$), opposite is $HG$. So cosine is adjacent over hypotenuse, so $5/13$. Wait, but let's confirm with the triangle. Alternatively, maybe I made a mistake. Wait, let's list the sides:

• Right angle at $G$, so $FG \perp HG$.

For $\angle F$:

• The two sides forming $\angle F$ are $FF$? No, $F$ is connected to $G$ and $H$. So $\angle F$ is between $FG$ and $FH$. So $FG$ is one leg (length 5), $FH$ is hypotenuse (13), and $HG$ is the other leg (12). So adjacent to $\angle F$ is $FG$ (length 5), hypotenuse is 13. So $\cos(\angle F) = \frac{FG}{FH} = \frac{5}{13}$.

Wait, but let's check the options. The options are 13/5, 5/13, 13/12, 12/13. So 5/13 is one of the options. So that's the answer.

Answer:

$\frac{5}{13}$ (corresponding to the option with $\frac{5}{13}$)