Question

find the exact value of cos y in simplest form.
triangle with right angle at x, sides xw=9, xy=√19, yw=10
answer attempt 1 out of 2
cos y = blank
submit answer

Explanation:

Step1: Recall cosine definition

In a right triangle, $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$. For $\angle Y$, adjacent side is $XY = \sqrt{19}$, hypotenuse is $WY = 10$. Wait, no, wait: Wait, triangle $XWY$ is right-angled at $X$. So sides: $XY = \sqrt{19}$, $XW = 9$, $WY = 10$ (since by Pythagoras, $(\sqrt{19})^2 + 9^2 = 19 + 81 = 100 = 10^2$, so $WY = 10$). So for angle $Y$, the adjacent side (to angle $Y$) is $XY$, and the hypotenuse is $WY$? Wait, no: in right triangle at $X$, angle at $Y$: the sides: opposite to $Y$ is $XW = 9$, adjacent to $Y$ is $XY = \sqrt{19}$, hypotenuse is $WY = 10$. Wait, no, cosine of angle $Y$: in triangle $XYW$, right-angled at $X$, so angle at $Y$: the adjacent side is $XY$, hypotenuse is $WY$. Wait, no, let's label the triangle: vertices $X$ (right angle), $Y$, $W$. So sides: $XY$ (leg), $XW$ (leg), $WY$ (hypotenuse). So angle at $Y$: the sides forming angle $Y$ are $XY$ and $WY$, and the other side is $XW$. So for angle $Y$, adjacent side is $XY$, hypotenuse is $WY$. Wait, no, cosine is adjacent over hypotenuse. So adjacent to angle $Y$ is $XY$, hypotenuse is $WY$. Wait, $XY = \sqrt{19}$, $WY = 10$. Wait, but let's confirm: in right triangle, cosine of an acute angle is adjacent leg over hypotenuse. So angle $Y$: the legs are $XY$ (length $\sqrt{19}$) and $XW$ (length 9), hypotenuse $WY$ (length 10). So adjacent to angle $Y$ is $XY$, opposite is $XW$. So $\cos Y = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{XY}{WY} = \frac{\sqrt{19}}{10}$? Wait, no, wait: wait, maybe I mixed up. Wait, let's draw the triangle: $X$ is right angle, so $X$ is between $Y$ and $W$? No, the triangle is $X$, $Y$, $W$ with right angle at $X$. So $X$ is connected to $Y$ and $X$ is connected to $W$, and $Y$ is connected to $W$. So sides: $XY$ (from $X$ to $Y$), $XW$ (from $X$ to $W$), $YW$ (from $Y$ to $W$). So angle at $Y$: between $XY$ and $YW$. So the adjacent side to angle $Y$ is $XY$, and the hypotenuse is $YW$. So yes, $\cos Y = \frac{XY}{YW} = \frac{\sqrt{19}}{10}$? Wait, but let's check Pythagoras: $XY^2 + XW^2 = (\sqrt{19})^2 + 9^2 = 19 + 81 = 100 = 10^2 = YW^2$, so that's correct. So $\cos Y = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{XY}{YW} = \frac{\sqrt{19}}{10}$. Wait, no, wait: wait, maybe I got adjacent and opposite wrong. Let's think again. In angle $Y$, the sides: the two sides forming angle $Y$ are $YX$ (length $\sqrt{19}$) and $YW$ (length 10), and the side opposite angle $Y$ is $XW$ (length 9). So in trigonometry, for angle $Y$, $\cos Y = \frac{\text{adjacent side}}{\text{hypotenuse}}$. The adjacent side is the one that is part of angle $Y$ and is not the hypotenuse. So the two sides of angle $Y$ are $YX$ (length $\sqrt{19}$) and $YW$ (hypotenuse, length 10). So the adjacent side is $YX$, hypotenuse is $YW$. So $\cos Y = \frac{YX}{YW} = \frac{\sqrt{19}}{10}$. Wait, but let's confirm with SOHCAHTOA: SOH (sine = opposite/hypotenuse), CAH (cosine = adjacent/hypotenuse), TOA (tangent = opposite/adjacent). So for angle $Y$, opposite side is $XW = 9$, adjacent side is $XY = \sqrt{19}$, hypotenuse is $YW = 10$. So cosine is adjacent over hypotenuse, so $\cos Y = \frac{\sqrt{19}}{10}$. Wait, but that seems small. Wait, maybe I made a mistake in identifying the sides. Wait, maybe the right angle is at $X$, so $X$ is the right angle, so $XY$ and $XW$ are the legs, $YW$ is the hypotenuse. So angle at $Y$: between $XY$ and $YW$. So the adjacent side to angle $Y$ is $XY$, and the hypotenuse is $YW$. So yes, $\cos Y = \frac{XY}{YW} = \frac{\sqrt{19}}{10}$. Wait, but let's check the lengths again. $XY = \sqrt{19}$, $XW = 9$, so $YW = \sqrt{(\sqrt{19})^2 + 9^2} = \sqrt{19 + 81} = \sqrt{100} = 10$. Correct. So $\cos Y = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{19}}{10}$. Wait, but maybe I mixed up adjacent and opposite. Wait, no: in angle $Y$, the side opposite is $XW$ (since it's not connected to $Y$), and the side adjacent is $XY$ (connected to $Y$ and the right angle). So yes, adjacent is $XY$, hypotenuse is $YW$. So $\cos Y = \frac{\sqrt{19}}{10}$.

Step2: Confirm the values

Adjacent side to $\angle Y$: $XY = \sqrt{19}$
Hypotenuse: $WY = 10$
Thus, $\cos Y = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{19}}{10}$

Answer:

$\frac{\sqrt{19}}{10}$