Question

solve for x. round to the nearest tenth of a degree, if necessary.

Explanation:

Step1: Identify trigonometric ratio

In right triangle \( \triangle QRS \) (right - angled at \( R \)), we know the length of the opposite side to angle \( x \) (which is \( RQ \)? Wait, no. Wait, angle at \( S \) is \( x^{\circ} \), the side adjacent to angle \( x \) is \( SR = 34 \), and the hypotenuse is \( SQ=48 \). Wait, no, in a right - triangle, for angle \( x \) at \( S \), the adjacent side is \( SR \) and the hypotenuse is \( SQ \)? Wait, no, \( \cos x=\frac{\text{adjacent}}{\text{hypotenuse}} \), \( \sin x = \frac{\text{opposite}}{\text{hypotenuse}} \), \( \tan x=\frac{\text{opposite}}{\text{adjacent}} \). Here, \( SR = 34 \) (opposite to angle \( Q \)), \( RQ \) (let's say) is adjacent to angle \( Q \), but angle at \( S \) is \( x \). So for angle \( x \) at \( S \), the opposite side is \( RQ \)? Wait, no, the right angle is at \( R \), so sides: \( SR = 34 \) (one leg), \( RQ \) (another leg), \( SQ = 48 \) (hypotenuse). So for angle \( x \) (at \( S \)), the adjacent side is \( SR = 34 \), and the hypotenuse is \( SQ = 48 \). So we can use the cosine function: \( \cos x=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{34}{48} \)

Step2: Solve for \( x \)

We have \( \cos x=\frac{34}{48}\approx0.7083 \)
To find \( x \), we take the inverse cosine (arccos) of \( 0.7083 \): \( x=\arccos(0.7083) \)
Using a calculator, \( x\approx44.9^{\circ} \) (rounded to the nearest tenth of a degree)

Answer:

\( 44.9^{\circ} \)