Question

solve for x. round to the nearest tenth, if necessary.
triangle image with right angle at n, angle at o is 44°, side on is 9, side om is x
answer attempt 2 out of 2
x = blank box

Explanation:

Step1: Identify trigonometric ratio

In right triangle \( ONM \), \( \angle N = 90^\circ \), \( \angle O = 44^\circ \), hypotenuse \( ON = 9 \)? Wait, no, \( ON \) is not hypotenuse. Wait, \( \angle N \) is right angle, so sides: \( ON \) is adjacent to \( \angle O \), \( MN \) is opposite, \( OM \) is hypotenuse? Wait, no, \( \angle N \) is right angle, so \( ON \) and \( MN \) are legs, \( OM \) is hypotenuse. Wait, \( \angle O = 44^\circ \), side \( ON \) is adjacent? Wait, no, \( x \) is \( OM \)? Wait, no, \( x \) is \( OM \)? Wait, the triangle: \( O \) at top, \( N \) at right (right angle), \( M \) at bottom. So \( ON \) is a leg (length 9), \( \angle O = 44^\circ \), \( x \) is the adjacent side? Wait, no, cosine of \( \angle O \) is adjacent over hypotenuse? Wait, no, \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \). Wait, \( \angle O = 44^\circ \), adjacent side to \( \angle O \) is \( x \)? No, wait, \( ON \) is length 9, which is opposite? Wait, no, let's label: \( \angle N = 90^\circ \), so \( \triangle ONM \) has right angle at \( N \), so sides: \( ON \) (from \( O \) to \( N \)), \( MN \) (from \( M \) to \( N \)), \( OM \) (from \( O \) to \( M \), hypotenuse). \( \angle O = 44^\circ \), so \( \cos(44^\circ) = \frac{\text{adjacent to } \angle O}{\text{hypotenuse}} = \frac{x}{OM} \)? No, wait, \( ON \) is length 9, which is the adjacent side? Wait, no, \( \angle O \) is at \( O \), so the sides: adjacent is \( OM \)? No, I'm confused. Wait, \( \angle N \) is right angle, so \( ON \) and \( MN \) are legs, \( OM \) is hypotenuse. \( \angle O = 44^\circ \), so the side opposite \( \angle O \) is \( MN \), adjacent is \( ON \) (length 9), and hypotenuse is \( OM = x \)? Wait, no, \( x \) is \( OM \)? Wait, the problem says \( x \) is \( OM \)? Wait, the diagram: \( O \) to \( M \) is \( x \), \( O \) to \( N \) is 9, \( N \) to \( M \) is leg. So \( \cos(44^\circ) = \frac{ON}{OM} \), because \( ON \) is adjacent to \( \angle O \), \( OM \) is hypotenuse. So \( \cos(44^\circ) = \frac{9}{x} \), so \( x = \frac{9}{\cos(44^\circ)} \).

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

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

Step3: Solve for \( x \)

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

Wait, wait, maybe I mixed up. Wait, \( \angle N \) is right angle, so \( ON \) is a leg (length 9), \( \angle O = 44^\circ \), 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, maybe \( \sin(44^\circ) \)? Wait, no, let's check again. In right triangle, \( \cos(\theta) = \text{adjacent}/\text{hypotenuse} \). \( \theta = 44^\circ \), adjacent side is \( ON \) (length 9), hypotenuse is \( OM \) (length \( x \)). So yes, \( \cos(44^\circ) = 9/x \), so \( x = 9 / \cos(44^\circ) \approx 9 / 0.7193 \approx 12.5 \).

Answer:

\( x \approx 12.5 \)