Question

solve for x. round to the nearest tenth of a degree, if necessary.
(image of a right triangle klm with right angle at l, kl = 69, km = 92, angle at m is x°)

Explanation:

Step1: Identify triangle type and trigonometric ratio

This is a right triangle with right angle at \( L \). We know the length of the adjacent side to angle \( x \) (which is \( LM \))? Wait, no, wait. Wait, in triangle \( KLM \), right-angled at \( L \), \( KL = 69 \), \( KM = 92 \), and we need to find angle \( x \) at \( M \). Wait, actually, for angle \( x \) at \( M \), the opposite side is \( KL = 69 \), and the hypotenuse is \( KM = 92 \). So we can use the sine function, since \( \sin(x) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{KL}{KM} \).

So \( \sin(x) = \frac{69}{92} \).

Step2: Solve for \( x \)

To find \( x \), we take the inverse sine (arcsin) of \( \frac{69}{92} \).

First, calculate \( \frac{69}{92} \). Simplify: \( \frac{69}{92} = \frac{3}{4} = 0.75 \).

Then, \( x = \arcsin(0.75) \).

Using a calculator, \( \arcsin(0.75) \approx 48.6^\circ \) (rounded to the nearest tenth).

Answer:

\( 48.6^\circ \)