Question

express cos u as a fraction in simplest terms.
(image of right triangle tus with right angle at t, tu = 10, us = 15)
answer
cos u = blank

Explanation:

Step1: Recall cosine definition

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

Step2: Identify sides for $\angle U$

For $\angle U$, adjacent side is $UT = 10$, hypotenuse is $US = 15$? Wait, no, wait. Wait, triangle $TUS$ is right-angled at $T$. So sides: $UT = 10$, $TS$ is another leg, $US = 15$ (hypotenuse? Wait, no, in right triangle, hypotenuse is opposite right angle. So right angle at $T$, so hypotenuse is $US$. Then for $\angle U$, adjacent side is $UT$, opposite is $TS$, hypotenuse is $US$. Wait, no: $\cos U = \frac{\text{adjacent to } U}{\text{hypotenuse}}$. Adjacent to $U$ is $UT$? Wait, no, let's label the triangle. Right angle at $T$, so vertices: $T$ (right angle), $U$, $S$. So sides: $UT = 10$, $TT$? No, $UT = 10$, $TS$ is unknown, $US = 15$ (the side opposite right angle $T$, so hypotenuse is $US = 15$? Wait, no, hypotenuse is the side opposite the right angle, so right angle at $T$, so hypotenuse is $US$. Then, for angle $U$, the adjacent side is $UT$, and the hypotenuse is $US$? Wait, no, adjacent side is the leg that forms angle $U$ with the hypotenuse. So angle $U$ is at vertex $U$, so the sides: $UT$ (one leg), $US$ (hypotenuse), and $TS$ (other leg). Wait, no, in triangle $UTS$, right-angled at $T$, so:

- $\angle T = 90^\circ$
- $UT = 10$ (leg)
- $US = 15$ (hypotenuse? Wait, no, hypotenuse should be the longest side. Wait, maybe I made a mistake. Wait, if it's a right triangle at $T$, then the sides are $UT$, $TS$, and $US$ (hypotenuse). So by Pythagoras, $UT^2 + TS^2 = US^2$. But $UT = 10$, $US = 15$, so $TS^2 = 15^2 - 10^2 = 225 - 100 = 125$, so $TS = 5\sqrt{5}$. But for $\cos U$, we need adjacent over hypotenuse. Adjacent to $U$ is $UT$, and hypotenuse is $US$? Wait, no, adjacent to angle $U$ is the leg that is part of angle $U$ and not the hypotenuse. So angle $U$ is between $UT$ and $US$? No, angle $U$ is at vertex $U$, so the two sides forming angle $U$ are $UT$ (from $U$ to $T$) and $US$ (from $U$ to $S$). Wait, no, in triangle, angle at $U$ is between $UU$? No, vertices are $U$, $T$, $S$. So angle at $U$ is between $UT$ (segment from $U$ to $T$) and $US$ (segment from $U$ to $S$). So the adjacent side to angle $U$ is $UT$, and the hypotenuse is $US$? Wait, no, hypotenuse is opposite the right angle, so right angle at $T$, so hypotenuse is $US$. Then, $\cos U = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{UT}{US}$? Wait, no, that can't be, because $UT = 10$, $US = 15$, but then $\cos U = 10/15 = 2/3$, but wait, maybe I mixed up adjacent and opposite. Wait, no, in cosine, it's adjacent over hypotenuse. Let's recall: in a right triangle, for an acute angle $\theta$:

- $\cos \theta = \frac{\text{length of adjacent side to } \theta}{\text{length of hypotenuse}}$
- $\sin \theta = \frac{\text{length of opposite side to } \theta}{\text{length of hypotenuse}}$
- $\tan \theta = \frac{\text{length of opposite side to } \theta}{\text{length of adjacent side to } \theta}$

So angle $U$: let's identify the sides. The right angle is at $T$, so:

- Opposite to angle $U$: $TS$ (because it's across from $U$)
- Adjacent to angle $U$: $UT$ (because it's next to $U$ and forms the angle with the hypotenuse)
- Hypotenuse: $US$ (opposite right angle $T$)

Wait, but then $\cos U = \frac{UT}{US} = \frac{10}{15} = \frac{2}{3}$? But wait, that seems too simple. Wait, maybe I got the hypotenuse wrong. Wait, if the right angle is at $T$, then the hypotenuse is $US$, so $US$ is the hypotenuse. Then $UT = 10$ (adjacent to $U$), $US = 15$ (hypotenuse). So $\cos U = \frac{UT}{US} = \frac{10}{15} = \frac{2}{3}$. Wait, but let's check with Pythagoras. If $UT = 10$, $US = 15$, then $TS = \sqrt{15^2 - 10^2} = \sqrt{225 - 100} = \sqrt{125} = 5\sqrt{5}$, which is correct. So then $\cos U = \frac{adjacent}{hypotenuse} = \frac{UT}{US} = \frac{10}{15} = \frac{2}{3}$. Wait, but maybe I mixed up adjacent and hypotenuse. Wait, no, adjacent is the leg that is part of the angle, so angle $U$ is between $UT$ and $US$, so $UT$ is adjacent, $US$ is hypotenuse. So yes, $\cos U = \frac{10}{15} = \frac{2}{3}$.

Wait, but let's confirm the definition again. In a right triangle, cosine of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse. So for angle $U$, the adjacent side is $UT$ (since it's one of the sides forming angle $U$ and is not the hypotenuse), and the hypotenuse is $US$ (the side opposite the right angle). Therefore, $\cos U = \frac{UT}{US} = \frac{10}{15} = \frac{2}{3}$.

Answer:

$\frac{2}{3}$