Question

find the ratio for cos ∠c

Explanation:

Step1: Recall cosine definition

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

Step2: Identify sides for $\angle C$

For $\angle C$, the adjacent side (to $\angle C$) is $AC = 7$, and the hypotenuse is $CT = 25$? Wait, no, wait. Wait, the right angle is at $A$, so triangle $ACT$ is right-angled at $A$. So sides: $AC = 7$, $AT = 24$, $CT = 25$ (since $7^2 + 24^2 = 49 + 576 = 625 = 25^2$). So for $\angle C$, the adjacent side is $AC = 7$? No, wait, adjacent side to $\angle C$: in right triangle at $A$, angle at $C$: the sides: adjacent is $AC$? Wait, no, let's label the triangle: right angle at $A$, so vertices $A$, $C$, $T$ with $A$ right angle. So $AC$ and $AT$ are legs, $CT$ is hypotenuse. So angle at $C$: the sides: adjacent to $\angle C$ is $AC$ (length 7), opposite is $AT$ (length 24), hypotenuse is $CT$ (length 25). Wait, no, wait: $\cos(\angle C) = \frac{\text{adjacent to } \angle C}{\text{hypotenuse}}$. Adjacent to $\angle C$ is the side that is part of $\angle C$ and not the hypotenuse. So $\angle C$ is between $AC$ and $CT$, so adjacent side is $AC = 7$, hypotenuse is $CT = 25$? Wait, no, that can't be, because 7/25 is an option (red triangle with 7/25). Wait, let's recheck. Wait, maybe I mixed up the sides. Wait, in right triangle $ACT$, right-angled at $A$, so:

• $\angle A = 90^\circ$
• $AC = 7$ (one leg)
• $AT = 24$ (another leg)
• $CT = 25$ (hypotenuse, since $7^2 + 24^2 = 25^2$)

Now, for $\angle C$:

• The sides:
• Adjacent side: the side that is adjacent to $\angle C$ and is a leg (not hypotenuse). So $\angle C$ is formed by $AC$ and $CT$. So the adjacent side is $AC = 7$, and the hypotenuse is $CT = 25$. Wait, but then $\cos(\angle C) = \frac{AC}{CT} = \frac{7}{25}$? But wait, another way: maybe I got the angle wrong. Wait, no, let's check the options. One of the options is 7/25 (red triangle). Wait, but let's confirm the cosine ratio. Wait, maybe I made a mistake. Wait, in a right triangle, cosine of an angle is adjacent over hypotenuse. So for angle at $C$:

Adjacent side: the side that is next to angle $C$ and is not the hypotenuse. So angle $C$ is between $AC$ and $CT$, so adjacent is $AC = 7$, hypotenuse is $CT = 25$. So $\cos(\angle C) = \frac{AC}{CT} = \frac{7}{25}$. Wait, but let's check the other sides. Wait, maybe the adjacent side is $AC$? Wait, no, wait, maybe I mixed up the angle. Wait, maybe the angle is at $T$? No, the problem says $\cos \angle C$. So $\angle C$ is at vertex $C$. So yes, adjacent side is $AC = 7$, hypotenuse is $CT = 25$. So $\cos(\angle C) = \frac{7}{25}$.

Wait, but let's check the options. The red triangle has 7/25, which matches. So that's the ratio.

Answer:

$\frac{7}{25}$ (corresponding to the option with 7/25, e.g., the red triangle option)