To find the total displacement, we need to break down each movement into its East-West (x) and North-South (y) components, then sum the components, and finally calculate the magnitude of the resultant vector. Let East be the positive x-direction and North be the positive y-direction. Step 1: Resolve the first displacement into components. The man moves $6.0 \text{ m}$ East. $$D_1 = (6.0 \text{ m}, 0 \text{ m})$$ So, $D_{1x} = 6.0 \text{ m}$ and $D_{1y} = 0 \text{ m}$. Step 2: Resolve the second displacement into components. The man moves $10.0 \text{ m}$ in the direction N30°E. This means $30^\circ$ East of North. The angle with respect to the positive y-axis (North) is $30^\circ$. The x-component (East) is $D_2 \sin(30^\circ)$. The y-component (North) is $D_2 \cos(30^\circ)$. $$D_{2x} = 10.0 \text{ m} \times \sin(30^\circ)$$ $$D_{2y} = 10.0 \text{ m} \times \cos(30^\circ)$$ We know that $\sin(30^\circ) = 0.5$ and $\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$. $$D_{2x} = 10.0 \text{ m} \times 0.5 = 5.0 \text{ m}$$ $$D_{2y} = 10.0 \text{ m} \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \text{ m} \approx 8.66 \text{ m}$$ Step 3: Sum the x and y components to find the total displacement. Total x-displacement ($R_x$): $$R_x = D_{1x} + D_{2x} = 6.0 \text{ m} + 5.0 \text{ m} = 11.0 \text{ m}$$ Total y-displacement ($R_y$): $$R_y = D_{1y} + D_{2y} = 0 \text{ m} + 5\sqrt{3} \text{ m} = 5\sqrt{3} \text{ m}$$ Step 4: Calculate the magnitude of the resultant displacement. The magnitude of the resultant displacement $R$ is given by the Pythagorean theorem: $$R = \sqrt{R_x^2 + R_y^2}$$ $$R = \sqrt{(11.0 \text{ m})^2 + (5\sqrt{3} \text{ m})^2}$$ $$R = \sqrt{121 \text{ m}^2 + (25 \times 3) \text{ m}^2}$$ $$R = \sqrt{121 \text{ m}^2 + 75 \text{ m}^2}$$ $$R = \sqrt{196 \text{ m}^2}$$ $$R = 14.0 \text{ m}$$ The man is $14.0 \text{ m}$ from his starting point. Comparing this with the given options: A. 12.0 m B. 17.0 m C. 15.0 m D. 14.0 m The correct option is D. The final answer is $\boxed{\text{D. 14.0 m}}$.