Question

solve for x. round to the nearest tenth, if necessary. triangle image with right angle at l, angle at m is 31°, side ml is x, side mk is 96

Explanation:

Step1: Identify trigonometric ratio

In right triangle \( \triangle MLK \) (right-angled at \( L \)), we know angle \( M = 31^\circ \), adjacent side to \( \angle M \) is \( x \), and hypotenuse is \( 96 \). Wait, no—wait, adjacent? Wait, \( \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}} \)? Wait, no, \( \cos(31^\circ)=\frac{x}{96} \)? Wait, no, wait: \( \angle M = 31^\circ \), side \( ML = x \) (adjacent to \( \angle M \)), side \( MK = 96 \) (hypotenuse). So \( \cos(31^\circ)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{x}{96} \).

Step2: Solve for \( x \)

Rearrange the formula: \( x = 96 \times \cos(31^\circ) \). Calculate \( \cos(31^\circ) \approx 0.8572 \). Then \( x = 96 \times 0.8572 \approx 82.3 \). Wait, no—wait, maybe I mixed up. Wait, maybe it's \( \tan \)? Wait, no, right angle at \( L \), so sides: \( ML = x \) (one leg), \( MK = 96 \) (hypotenuse), \( LK \) (other leg). Angle at \( M \) is \( 31^\circ \), so \( \cos(31^\circ)=\frac{ML}{MK}=\frac{x}{96} \), so \( x = 96 \cos(31^\circ) \). Let's compute: \( \cos(31^\circ) \approx 0.857167 \), so \( 96 \times 0.857167 \approx 82.3 \). Wait, but maybe I made a mistake. Wait, alternatively, if \( x \) is the adjacent, hypotenuse is 96, angle 31, so cosine. Yes.

Wait, let's check with calculator: \( \cos(31^\circ) \approx 0.8572 \), \( 96 \times 0.8572 = 96 \times 0.8572 \). Let's compute: 960.8 = 76.8, 960.0572=5.4912, total 76.8+5.4912=82.2912, which rounds to 82.3.

Answer:

\( x \approx 82.3 \)