1. **State the problem:** We have a right triangle with hypotenuse $AB = 18$ and one leg $AC = 9\sqrt{3}$. We need to find the other leg $BC$ and the angles $\angle B$ and $\angle A$. 2. **Use the Pythagorean theorem:** In a right triangle, $AB^2 = AC^2 + BC^2$. 3. **Calculate $BC$:** $$ BC^2 = AB^2 - AC^2 = 18^2 - (9\sqrt{3})^2 = 324 - 81 \times 3 = 324 - 243 = 81 $$ $$ BC = \sqrt{81} = 9 $$ 4. **Find angles using trigonometry:** - Use sine or cosine. For example, $\sin \angle B = \frac{AC}{AB} = \frac{9\sqrt{3}}{18} = \frac{\cancel{9}\sqrt{3}}{\cancel{18}2} = \frac{\sqrt{3}}{2}$. 5. **Calculate $\angle B$:** $$ \angle B = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = 60^\circ $$ 6. **Calculate $\angle A$:** Since the triangle is right angled, $$ \angle A = 90^\circ - \angle B = 90^\circ - 60^\circ = 30^\circ $$ **Final answers:** - $BC = 9$ - $m\angle B = 60^\circ$ - $m\angle A = 30^\circ$