1. **State the problem:** We have a right triangle with vertices A, B, and C, where $\angle C = 90^\circ$. The lengths of sides are $AB = 9$ units and $AC = 7$ units. We need to find the measure of $\angle B$ rounded to the nearest hundredth. 2. **Identify the sides relative to $\angle B$:** - Side opposite $\angle B$ is $AC = 7$. - Side adjacent to $\angle B$ is $BC$ (unknown). - Hypotenuse is $AB = 9$. 3. **Use the Pythagorean theorem to find $BC$:** $$BC = \sqrt{AB^2 - AC^2} = \sqrt{9^2 - 7^2} = \sqrt{81 - 49} = \sqrt{32} = 4\sqrt{2}$$ 4. **Use the tangent function to find $\angle B$:** $$\tan(\angle B) = \frac{\text{opposite}}{\text{adjacent}} = \frac{AC}{BC} = \frac{7}{4\sqrt{2}}$$ 5. **Simplify the fraction:** $$\frac{7}{4\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{7\sqrt{2}}{4 \times 2} = \frac{7\sqrt{2}}{8}$$ 6. **Calculate $\angle B$ using arctangent:** $$\angle B = \tan^{-1}\left(\frac{7\sqrt{2}}{8}\right)$$ 7. **Evaluate the value:** $$\frac{7\sqrt{2}}{8} \approx \frac{7 \times 1.4142}{8} = \frac{9.8994}{8} = 1.2374$$ $$\angle B \approx \tan^{-1}(1.2374) \approx 51.34^\circ$$ **Final answer:** $\angle B \approx 51.34^\circ$ rounded to the nearest hundredth.