1. **Problem Statement:** Given a right triangle with a base of 16, a vertical side labeled $n$, a hypotenuse labeled $m$, and an angle of $30^\circ$ opposite side $n$, find the values of $m$ and $n$. 2. **Formula and Rules:** In a right triangle, the side opposite the $30^\circ$ angle is half the hypotenuse. Also, the side adjacent to the $30^\circ$ angle can be found using the Pythagorean theorem or trigonometric ratios. 3. **Step 1: Identify sides relative to the $30^\circ$ angle:** - Opposite side to $30^\circ$ is $n$ - Hypotenuse is $m$ - Adjacent side (base) is 16 4. **Step 2: Use the property of a $30^\circ$ right triangle:** $$n = \frac{m}{2}$$ 5. **Step 3: Use Pythagorean theorem:** $$m^2 = n^2 + 16^2$$ 6. **Step 4: Substitute $n = \frac{m}{2}$ into the Pythagorean theorem:** $$m^2 = \left(\frac{m}{2}\right)^2 + 16^2$$ $$m^2 = \frac{m^2}{4} + 256$$ 7. **Step 5: Simplify and solve for $m^2$:** $$m^2 - \frac{m^2}{4} = 256$$ $$\frac{4m^2}{4} - \frac{m^2}{4} = 256$$ $$\frac{3m^2}{4} = 256$$ 8. **Step 6: Multiply both sides by $\frac{4}{3}$ to isolate $m^2$:** $$m^2 = 256 \times \frac{4}{3}$$ $$m^2 = \frac{1024}{3}$$ 9. **Step 7: Take the square root of both sides:** $$m = \sqrt{\frac{1024}{3}} = \frac{32}{\sqrt{3}} = \frac{32\sqrt{3}}{3}$$ 10. **Step 8: Calculate $n$ using $n = \frac{m}{2}$:** $$n = \frac{1}{2} \times \frac{32\sqrt{3}}{3} = \frac{16\sqrt{3}}{3}$$ 11. **Step 9: Approximate values for clarity:** $$m \approx 18.48$$ $$n \approx 9.24$$ 12. **Step 10: Check given options:** - Option A: $m=24$, $n=8$ (No) - Option B: $m=8\sqrt{3} \approx 13.86$, $n=4$ (No) - Option C: $m=8\sqrt{3} \approx 13.86$, $n=8$ (No) - Option D: $m=24$, $n=4$ (No) None of the options exactly match the calculated values, but if the base is 16 and the angle is $30^\circ$, the side opposite should be $8$ (half the hypotenuse), so the hypotenuse should be $16$. Since the base is 16, this suggests the triangle is not drawn to scale or the base is adjacent to the $30^\circ$ angle. 13. **Alternative approach using trigonometry:** Since the base is adjacent to $30^\circ$, use cosine: $$\cos 30^\circ = \frac{16}{m}$$ $$\frac{\sqrt{3}}{2} = \frac{16}{m}$$ $$m = \frac{16 \times 2}{\sqrt{3}} = \frac{32}{\sqrt{3}} = \frac{32\sqrt{3}}{3}$$ Then, $$n = m \sin 30^\circ = \frac{32\sqrt{3}}{3} \times \frac{1}{2} = \frac{16\sqrt{3}}{3}$$ This matches the previous calculation. **Final answer:** $$m = \frac{32\sqrt{3}}{3}, \quad n = \frac{16\sqrt{3}}{3}$$ None of the options exactly match, but the closest is option C if $8\sqrt{3} \approx 13.86$ and $8$ is close to $n$ value.