Question

solve for x. round to the nearest tenth, if necessary. triangle with right angle at n, angle at o is 44°, side on is 9, side om is x

Explanation:

Step1: Identify trigonometric ratio

In right triangle \( ONM \), \( \angle O = 44^\circ \), opposite side to \( \angle O \) is \( 9 \), adjacent side is \( x \)? Wait, no: wait, \( \angle N \) is right angle, so \( ON \) is hypotenuse? Wait, no, \( ON \) is a side? Wait, no, the sides: \( \angle O = 44^\circ \), \( \angle N = 90^\circ \), so \( \sin(44^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} \)? Wait, no, \( \angle O \): the side opposite \( \angle O \) is \( MN \)? Wait, no, the triangle has vertices \( O \), \( N \), \( M \), with right angle at \( N \). So \( ON \) is one leg? Wait, no, \( ON \) is length 9? Wait, the side from \( O \) to \( N \) is length 9? Wait, no, the side labeled 9 is \( ON \)? Wait, no, the side from \( O \) to \( N \) is length 9, and \( \angle O = 44^\circ \), right angle at \( N \), so \( \cos(44^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} \)? Wait, no, \( x \) is \( OM \)? Wait, no, \( x \) is \( OM \)? Wait, no, \( x \) is \( OM \)? Wait, no, \( x \) is the side \( OM \)? Wait, no, \( x \) is \( OM \)? Wait, no, the side \( OM \) is \( x \), and \( ON \) is 9, with \( \angle O = 44^\circ \), right angle at \( N \). So in triangle \( ONM \), right-angled at \( N \), \( \cos(44^\circ) = \frac{ON}{OM} \), because \( ON \) is adjacent to \( \angle O \), and \( OM \) is the hypotenuse? Wait, no: \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \). So \( \theta = 44^\circ \), adjacent side is \( ON = 9 \), hypotenuse is \( OM = x \)? Wait, no, that can't be. Wait, maybe \( \tan(44^\circ) = \frac{MN}{ON} \), but \( x \) is \( OM \). Wait, no, let's re-express:

In right triangle \( ONM \), \( \angle N = 90^\circ \), \( \angle O = 44^\circ \), side \( ON = 9 \) (adjacent to \( \angle O \)), and \( OM = x \) (hypotenuse). So \( \cos(44^\circ) = \frac{ON}{OM} \), so \( \cos(44^\circ) = \frac{9}{x} \)? Wait, no, that would mean \( x = \frac{9}{\cos(44^\circ)} \). Wait, no, maybe I got the sides wrong. Wait, the side labeled 9 is \( ON \), and \( x \) is \( OM \). So \( \cos(44^\circ) = \frac{ON}{OM} \), so \( x = \frac{9}{\cos(44^\circ)} \). Wait, no, \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \), so adjacent is \( ON = 9 \), hypotenuse is \( OM = x \), so \( \cos(44^\circ) = \frac{9}{x} \), so \( x = \frac{9}{\cos(44^\circ)} \).

Wait, alternatively, maybe \( \sin(44^\circ) = \frac{MN}{OM} \), but \( MN \) is not labeled. Wait, no, the side labeled 9 is \( ON \), and \( x \) is \( OM \). Let's check the trigonometric ratios.

Wait, \( \angle O = 44^\circ \), right angle at \( N \), so:

\( \cos(44^\circ) = \frac{\text{adjacent to } \angle O}{\text{hypotenuse}} = \frac{ON}{OM} \)

So \( ON = 9 \), \( OM = x \), so:

\( \cos(44^\circ) = \frac{9}{x} \)

Solving for \( x \):

\( x = \frac{9}{\cos(44^\circ)} \)

Step2: Calculate \( \cos(44^\circ) \)

Using a calculator, \( \cos(44^\circ) \approx 0.7193 \)

Step3: Compute \( x \)

\( x = \frac{9}{0.7193} \approx 12.5 \) (rounded to the nearest tenth)

Wait, but wait, maybe I mixed up the sides. Let's re-examine the triangle. The right angle is at \( N \), so the sides are:

- \( ON \): length 9 (adjacent to \( \angle O \))
- \( MN \): opposite to \( \angle O \)
- \( OM \): hypotenuse (length \( x \))

So \( \cos(44^\circ) = \frac{ON}{OM} \), so \( OM = \frac{ON}{\cos(44^\circ)} = \frac{9}{\cos(44^\circ)} \approx \frac{9}{0.7193} \approx 12.5 \)

Alternatively, if \( x \) is the adjacent side, but that doesn't make sense. Wait, maybe I made a mistake. Let's check again.

Wait, maybe \( \angle O = 44^\circ \), and the side opposite is 9, and \( x \) is the adjacent side. Then \( \tan(44^\circ) = \frac{9}{x} \), so \( x = \frac{9}{\tan(44^\circ)} \). Let's check that. \( \tan(44^\circ) \approx 0.9657 \), so \( x = \frac{9}{0.9657} \approx 9.3 \). But that contradicts the previous. So which is correct?

Wait, the triangle: vertices \( O \), \( N \), \( M \), right angle at \( N \). So:

- \( \angle O = 44^\circ \)
- \( \angle N = 90^\circ \)
- \( \angle M = 46^\circ \) (since angles sum to 180)

Sides:

- \( ON \): from \( O \) to \( N \) (length 9)
- \( MN \): from \( M \) to \( N \) (opposite \( \angle O \))
- \( OM \): from \( O \) to \( M \) (hypotenuse, length \( x \))

So \( \cos(\angle O) = \frac{ON}{OM} \), so \( \cos(44^\circ) = \frac{9}{x} \), so \( x = \frac{9}{\cos(44^\circ)} \approx \frac{9}{0.7193} \approx 12.5 \)

Alternatively, if \( ON \) is the opposite side, then \( \sin(44^\circ) = \frac{ON}{OM} \), so \( x = \frac{9}{\sin(44^\circ)} \approx \frac{9}{0.6947} \approx 12.95 \approx 13.0 \). Wait, this is confusing. Let's look at the labels again. The side labeled 9 is \( ON \), and \( x \) is \( OM \). Wait, maybe the right angle is at \( N \), so \( ON \) and \( MN \) are legs, \( OM \) is hypotenuse. Then \( \angle O = 44^\circ \), so \( \cos(44^\circ) = \frac{ON}{OM} \), so \( OM = \frac{ON}{\cos(44^\circ)} \). So \( ON = 9 \), so \( x = \frac{9}{\cos(44^\circ)} \approx 12.5 \) (rounded to nearest tenth).

Wait, let's calculate \( \cos(44^\circ) \):

\( \cos(44^\circ) \approx 0.7193 \)

\( x = \frac{9}{0.7193} \approx 12.5 \)

Yes, that seems right.

Answer:

\( \boxed{12.5} \)