Question

solve for x. round to the nearest tenth of a degree, if necessary.
triangle with right angle at r, q to r is 64, q to s is 71, angle at s is x°

Explanation:

Step1: Identify trigonometric ratio

In right triangle \( QRS \), \( \angle R = 90^\circ \), hypotenuse \( QS = 71 \), adjacent side to \( \angle S \) (which is \( x^\circ \)) is \( RS \), and opposite side is \( QR = 64 \). Wait, actually, for \( \angle S \), the adjacent side is \( RS \), opposite is \( QR \), hypotenuse \( QS \). Wait, no, let's correct: in \( \triangle QRS \), right-angled at \( R \), so \( \sin(x) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{QR}{QS} \), since \( QR \) is opposite \( \angle S \), and \( QS \) is hypotenuse. So \( \sin(x) = \frac{64}{71} \).

Step2: Calculate \( x \)

First, compute \( \frac{64}{71} \approx 0.9014 \). Then, \( x = \arcsin(0.9014) \). Using a calculator, \( \arcsin(0.9014) \approx 64.3^\circ \) (rounded to nearest tenth). Wait, wait, no: wait, \( QR \) is 64, \( QS \) is 71. Wait, maybe I mixed up. Wait, \( \angle Q \) and \( \angle S \): let's check again. \( \angle R = 90^\circ \), so \( \cos(x) = \frac{\text{adjacent to } x}{\text{hypotenuse}} \). Wait, \( \angle S \) is at vertex \( S \), so the sides: \( RS \) is adjacent to \( \angle S \), \( QR \) is opposite to \( \angle S \), hypotenuse \( QS = 71 \). Wait, no, \( QR \) is length 64, which is opposite \( \angle S \), so \( \sin(x) = \frac{QR}{QS} = \frac{64}{71} \). Let's compute that: \( 64 \div 71 \approx 0.9014 \). Then \( x = \arcsin(0.9014) \). Let's calculate that. Using calculator, \( \arcsin(0.9014) \approx 64.3^\circ \)? Wait, no, wait, maybe I made a mistake. Wait, \( \cos(x) \): wait, maybe \( \angle Q \) is \( x \)? No, the diagram shows \( x^\circ \) at \( S \). Wait, let's re-express: in right triangle, \( \cos(\angle S) = \frac{RS}{QS} \), but we don't know \( RS \). Wait, no, \( QR = 64 \), \( QS = 71 \), so by Pythagoras, \( RS = \sqrt{71^2 - 64^2} = \sqrt{5041 - 4096} = \sqrt{945} \approx 30.74 \). Then \( \cos(x) = \frac{RS}{QS} = \frac{30.74}{71} \approx 0.433 \), but that would give \( x \approx 64.3^\circ \) if we use \( \sin(x) = 64/71 \approx 0.901 \), so \( \arcsin(0.901) \approx 64.3^\circ \). Wait, let's check with calculator: \( \sin(64.3^\circ) \approx \sin(64^\circ 18') \approx 0.901 \), yes. So that's correct.

Answer:

\( 64.3^\circ \)