1. **Problem statement:** A rescue boat travels from the harbor to Area A on a bearing of 070° for 40 km, then from Area A to Area B on a bearing of 150° for 55 km. We need to find: - The bearing of Area A from Area B. - How far east Area B is from Area A. - The direct distance from the harbor to Area B. 2. **Formulas and rules:** - Bearings are measured clockwise from north. - To find coordinates from bearings and distances, use: $$x = d \times \sin(\theta)$$ $$y = d \times \cos(\theta)$$ where $\theta$ is the bearing angle. - To find the bearing from one point to another, calculate the angle of the vector connecting them relative to north. - Distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is: $$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ 3. **Calculate coordinates:** - Harbor at origin: $(0,0)$ - Area A: $$x_A = 40 \times \sin(70^\circ) = 40 \times 0.9397 = 37.59$$ $$y_A = 40 \times \cos(70^\circ) = 40 \times 0.3420 = 13.68$$ - Area B from Area A: $$x_B = x_A + 55 \times \sin(150^\circ) = 37.59 + 55 \times 0.5 = 37.59 + 27.5 = 65.09$$ $$y_B = y_A + 55 \times \cos(150^\circ) = 13.68 + 55 \times (-0.8660) = 13.68 - 47.63 = -33.95$$ 4. **Bearing of Area A from Area B:** - Vector from B to A: $$\Delta x = x_A - x_B = 37.59 - 65.09 = -27.5$$ $$\Delta y = y_A - y_B = 13.68 - (-33.95) = 47.63$$ - Angle from north: $$\theta = \arctan\left(\frac{|\Delta x|}{\Delta y}\right) = \arctan\left(\frac{27.5}{47.63}\right) = 30^\circ$$ - Since $\Delta x$ is negative and $\Delta y$ positive, vector points northwest quadrant, so bearing is: $$360^\circ - 30^\circ = 330^\circ$$ 5. **East distance from Area A to Area B:** - East distance is difference in $x$ coordinates: $$x_B - x_A = 65.09 - 37.59 = 27.5 \text{ km}$$ 6. **Direct distance from harbor to Area B:** $$d = \sqrt{(65.09)^2 + (-33.95)^2} = \sqrt{4236.7 + 1152.2} = \sqrt{5388.9} = 73.43 \text{ km}$$ **Final answers:** - Bearing of Area A from Area B: $330^\circ$ - East distance from Area A to Area B: 27.5 km - Direct distance from harbor to Area B: 73.43 km