Question

express the trig ratios as fractions in simplest terms.
triangle image with right angle at m, sides: nm=41, om=√83, on=42
answer
sin o =
cos n =
sin o and cos n

Explanation:

Step1: Recall Sine Definition

In a right triangle, $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$. For $\sin O$, angle $O$ has opposite side $NM = 41$ and hypotenuse $ON = 42$? Wait, no, wait. Wait, triangle $OMN$ is right-angled at $M$. So sides: $OM = \sqrt{83}$, $NM = 41$, $ON = 42$? Wait, no, let's check: right angle at $M$, so legs are $OM$ and $NM$, hypotenuse $ON$. Wait, $ON = 42$, $NM = 41$, $OM = \sqrt{83}$. Let's confirm Pythagoras: $41^2 + (\sqrt{83})^2 = 1681 + 83 = 1764 = 42^2$. Yes, correct. So for angle $O$: opposite side to $O$ is $NM = 41$, hypotenuse is $ON = 42$? Wait, no: angle $O$ is at vertex $O$, so in triangle $OMN$, angle at $O$: the sides: adjacent is $OM = \sqrt{83}$, opposite is $NM = 41$, hypotenuse $ON = 42$. So $\sin O = \frac{\text{opposite to } O}{\text{hypotenuse}} = \frac{NM}{ON} = \frac{41}{42}$? Wait, no, wait: wait, $ON$ is 42, $NM$ is 41, $OM$ is $\sqrt{83}$. Wait, angle at $O$: the sides: opposite side is $NM$ (since from $O$, the side opposite is $NM$), hypotenuse is $ON$. So $\sin O = \frac{NM}{ON} = \frac{41}{42}$? Wait, no, wait, maybe I mixed up. Wait, let's label the triangle: right angle at $M$, so vertices $O$, $M$, $N$. So $OM$ and $NM$ are legs, $ON$ is hypotenuse. So angle at $O$: between $ON$ and $OM$. So opposite side to $O$ is $NM$ (length 41), adjacent is $OM$ (length $\sqrt{83}$), hypotenuse $ON$ (length 42). So $\sin O = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{NM}{ON} = \frac{41}{42}$.

Step2: Recall Cosine Definition for $\cos N$

$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$. For angle $N$, in triangle $OMN$, right-angled at $M$. Angle at $N$: adjacent side is $NM = 41$, hypotenuse is $ON = 42$? Wait, no: angle at $N$, adjacent side is $NM$? Wait, angle at $N$: between $ON$ and $NM$. So adjacent side is $NM = 41$, hypotenuse $ON = 42$, opposite side is $OM = \sqrt{83}$. Wait, $\cos N = \frac{\text{adjacent to } N}{\text{hypotenuse}} = \frac{NM}{ON} = \frac{41}{42}$? Wait, no, wait: angle at $N$, the adjacent side is $NM$? Wait, no, angle at $N$: the sides: adjacent is $NM$ (since from $N$, the side adjacent to angle $N$ is $NM$), hypotenuse $ON$. Wait, no, angle at $N$: the sides: adjacent is $NM$? Wait, maybe better to use angle $N$: in triangle $OMN$, angle at $N$, the adjacent side is $NM$ (length 41), hypotenuse $ON$ (length 42), so $\cos N = \frac{NM}{ON} = \frac{41}{42}$? Wait, but wait, let's check angle $N$: the opposite side is $OM = \sqrt{83}$, adjacent is $NM = 41$, hypotenuse $ON = 42$. So $\cos N = \frac{\text{adjacent to } N}{\text{hypotenuse}} = \frac{NM}{ON} = \frac{41}{42}$. Wait, but that's the same as $\sin O$? Wait, maybe I made a mistake. Wait, no: $\sin O$ is opposite over hypotenuse, which is $NM/ON = 41/42$. $\cos N$: in angle $N$, adjacent over hypotenuse: $NM/ON = 41/42$? Wait, but let's check the triangle again. Wait, angle $O$ and angle $N$: in a right triangle, $\sin O = \cos N$ because they are complementary angles (since angle $O + angle N = 90^\circ$), so $\sin O = \cos (90^\circ - O) = \cos N$. So that makes sense. So let's confirm:

For $\sin O$:
- Opposite to $O$: $NM = 41$
- Hypotenuse: $ON = 42$
- So $\sin O = \frac{41}{42}$

For $\cos N$:
- Adjacent to $N$: $NM = 41$ (since angle $N$ is at $N$, the side adjacent is $NM$, as the other leg is $OM$, which is opposite to $N$)
- Hypotenuse: $ON = 42$
- So $\cos N = \frac{41}{42}$

Wait, but let's check again. Wait, angle $N$: the sides: adjacent is $NM$ (length 41), hypotenuse $ON$ (length 42), so $\cos N = \frac{41}{42}$. And $\sin O$: opposite to $O$ is $NM$ (length 41), hypotenuse $ON$ (length 42), so $\sin O = \frac{41}{42}$. So both are $\frac{41}{42}$.

Step2: Wait, Wait, Maybe I Mixed Up Sides

Wait, no, wait: $OM$ is $\sqrt{83}$, $NM$ is 41, $ON$ is 42. So angle $O$: the sides: opposite is $NM = 41$, hypotenuse $ON = 42$, so $\sin O = 41/42$. Angle $N$: adjacent is $NM = 41$, hypotenuse $ON = 42$, so $\cos N = 41/42$. Yes, that's correct because $O$ and $N$ are complementary (sum to 90 degrees), so $\sin O = \cos N$.

Answer:

$\sin O = \frac{41}{42}$, $\cos N = \frac{41}{42}$

Wait, but the problem says "Express the trig ratios as fractions in simplest terms. $\sin O = \square$" and " $\cos N = \square$". So for $\sin O$, the answer is $\frac{41}{42}$, and for $\cos N$, also $\frac{41}{42}$.