Question

find the cosine of ∠h.
f---37---h
| |
| 35 |12
| |
g (right angle at g)
simplify your answer and write it as a proper fraction, improper fraction, or whole number.
cos(h) =
submit

Explanation:

Step1: Recall cosine definition

In a right triangle, cosine of an angle is adjacent over hypotenuse. For $\angle H$, adjacent side is $GH = 12$, hypotenuse is $FH = 37$? Wait, no, wait. Wait, triangle $FGH$ is right-angled at $G$. So sides: $GH = 12$, $FG = 35$, $FH = 37$ (since $12^2 + 35^2 = 144 + 1225 = 1369 = 37^2$). So for $\angle H$, the adjacent side to $\angle H$ is $GH$ (length 12), and the hypotenuse is $FH$ (length 37)? Wait, no, wait. Wait, in right triangle at $G$, angle at $H$: the sides adjacent to $\angle H$ is $GH$, opposite is $FG$, hypotenuse is $FH$. Wait, no, cosine of $\angle H$: adjacent side is $GH$ (since $\angle H$ is at vertex $H$, so the sides forming $\angle H$ are $FH$ and $GH$, with right angle at $G$. So adjacent side to $\angle H$ is $GH$ (length 12), hypotenuse is $FH$ (length 37)? Wait, no, wait, no. Wait, let's label the triangle: right angle at $G$, so vertices $F$, $G$, $H$ with $G$ right angle. So sides: $FG = 35$, $GH = 12$, $FH = 37$. So angle at $H$: the sides are $GH$ (adjacent), $FH$ (hypotenuse), and $FG$ (opposite). So cosine of $\angle H$ is adjacent over hypotenuse, which is $GH / FH$? Wait, no, wait, adjacent to $\angle H$: the side that is part of $\angle H$ and not the hypotenuse. So $\angle H$ is between $GH$ and $FH$. So adjacent side is $GH$ (length 12), hypotenuse is $FH$ (length 37)? Wait, no, that can't be. Wait, no, wait, in a right triangle, cosine of an angle is adjacent leg over hypotenuse. Wait, angle at $H$: the legs are $GH$ (vertical) and $FG$ (horizontal)? Wait, no, $G$ is right angle, so $FG$ and $GH$ are the legs, $FH$ is hypotenuse. So for $\angle H$, the adjacent leg is $GH$ (since it's one of the legs forming $\angle H$), and the hypotenuse is $FH$. Wait, but let's check: $\cos(\angle H) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{GH}{FH}$? Wait, $GH = 12$, $FH = 37$, so $\cos(H) = \frac{12}{37}$? Wait, no, wait, no, that's wrong. Wait, wait, no, I think I mixed up. Wait, angle at $H$: the sides adjacent to $\angle H$ is $GH$ (length 12), and the hypotenuse is $FH$ (length 37). Wait, but let's confirm with the right triangle. Wait, $GH = 12$, $FG = 35$, $FH = 37$. So $\cos(\angle H) = \frac{\text{adjacent to } H}{\text{hypotenuse}} = \frac{GH}{FH} = \frac{12}{37}$? Wait, no, wait, no, that's not right. Wait, no, wait, adjacent to $\angle H$: the side that is next to $\angle H$ and is a leg. Wait, $\angle H$ is at $H$, so the two sides meeting at $H$ are $FH$ (hypotenuse) and $GH$ (leg). The other leg is $FG$, opposite to $\angle H$. So yes, adjacent to $\angle H$ is $GH$ (length 12), hypotenuse is $FH$ (length 37). So $\cos(H) = \frac{12}{37}$? Wait, but let's check again. Wait, maybe I made a mistake. Wait, no, in right triangle, cosine of angle is adjacent over hypotenuse. So for angle at $H$, adjacent side is $GH$ (12), hypotenuse is $FH$ (37). So $\cos(H) = \frac{12}{37}$. Wait, but let's confirm with the sides. $12^2 + 35^2 = 144 + 1225 = 1369 = 37^2$, so that's correct. So adjacent to $\angle H$ is $GH = 12$, hypotenuse $FH = 37$. So cosine is $\frac{12}{37}$.

Step2: Apply cosine formula

$\cos(H) = \frac{\text{adjacent to } H}{\text{hypotenuse}} = \frac{GH}{FH} = \frac{12}{37}$

Answer:

$\frac{12}{37}$