1. **Problem statement:** A boy starts at point A and walks 0.4 km to point D, then turns right 90° and walks 0.3 km to point C. The angle between AB and AD is $\theta$, and the perpendicular distance from C to AB is $L$. We need to show that $$L = 0.4 \sin \theta - 0.3 \cos \theta.$$ 2. **Understanding the problem:** - $AD = 0.4$ km - $DC = 0.3$ km - Angle $DAB = \theta$ - $L$ is the perpendicular distance from $C$ to $AB$ 3. **Approach:** We will use trigonometry and vector components to express $L$ in terms of $\theta$. 4. **Step 1: Express coordinates relative to A** Assume $A$ is at the origin $(0,0)$. - Vector $\overrightarrow{AB}$ lies along the x-axis. - Vector $\overrightarrow{AD}$ makes an angle $\theta$ with $\overrightarrow{AB}$. Coordinates of $D$: $$D_x = 0.4 \cos \theta, \quad D_y = 0.4 \sin \theta.$$ 5. **Step 2: Coordinates of C** Since $C$ is reached by walking 0.3 km perpendicular to $AD$ (turning right 90°), the direction from $D$ to $C$ is perpendicular to $AD$. The unit vector along $AD$ is: $$\hat{u}_{AD} = (\cos \theta, \sin \theta).$$ Turning right 90° means the vector from $D$ to $C$ is: $$\vec{DC} = 0.3 (\sin \theta, -\cos \theta).$$ So coordinates of $C$ are: $$C_x = D_x + 0.3 \sin \theta = 0.4 \cos \theta + 0.3 \sin \theta,$$ $$C_y = D_y - 0.3 \cos \theta = 0.4 \sin \theta - 0.3 \cos \theta.$$ 6. **Step 3: Calculate perpendicular distance $L$ from $C$ to $AB$** Since $AB$ lies along the x-axis, the perpendicular distance from $C$ to $AB$ is simply the absolute value of the y-coordinate of $C$: $$L = |C_y| = |0.4 \sin \theta - 0.3 \cos \theta|.$$ Given $0^\circ \leq \theta \leq 90^\circ$, the expression inside is positive, so: $$L = 0.4 \sin \theta - 0.3 \cos \theta.$$ **Final answer:** $$\boxed{L = 0.4 \sin \theta - 0.3 \cos \theta}.$$