1. **State the problem:** We have a right triangle P Q O with a right angle at P. Side QP is 73, side QO is 97, and we want to find the angle $x^\circ$ at vertex Q. 2. **Identify the sides relative to angle $x$:** - Side QP (73) is opposite angle $x$. - Side QO (97) is the hypotenuse. 3. **Use the sine function:** The sine of an angle in a right triangle is the ratio of the length of the opposite side to the hypotenuse. $$\sin(x) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{73}{97}$$ 4. **Calculate $x$ by taking the inverse sine:** $$x = \sin^{-1}\left(\frac{73}{97}\right)$$ 5. **Evaluate the fraction:** $$\frac{73}{97} \approx 0.7526$$ 6. **Find the angle:** $$x = \sin^{-1}(0.7526) \approx 48.7^\circ$$ 7. **Round to the nearest tenth:** $$x \approx 48.7^\circ$$ **Final answer:** $x = 48.7^\circ$