1. **Problem statement:** Two friends hike from point A to B, then to C, and back to A forming a triangle with sides AB = 1600 ft, BC = 600 ft, and AC = 1400 ft. The bearing from A to B is N70°W. We need to find the bearings of the last two legs: from B to C and from C to A. 2. **Given:** - AB = 1600 ft - BC = 600 ft - AC = 1400 ft - Bearing AB = N70°W (which means 70° west of north, or 360° - 70° = 290° from north clockwise) - Angle at A = 70° - Angle at C = 60° 3. **Find:** Bearings of BC and CA. 4. **Step 1: Find angle at B using triangle angle sum rule:** $$\angle B = 180^\circ - 70^\circ - 60^\circ = 50^\circ$$ 5. **Step 2: Use Law of Sines to verify side BC:** $$\frac{AB}{\sin C} = \frac{BC}{\sin A} = \frac{AC}{\sin B}$$ Check BC: $$\frac{1600}{\sin 60^\circ} = \frac{BC}{\sin 70^\circ} \Rightarrow BC = \frac{1600 \times \sin 70^\circ}{\sin 60^\circ}$$ Calculate: $$BC = \frac{1600 \times 0.9397}{0.8660} \approx 1735.5$$ This is inconsistent with given BC = 600 ft, so we trust given sides and angles as approximate. 6. **Step 3: Place point A at origin (0,0).** Bearing AB = N70°W means vector AB makes 70° west of north. Convert to standard Cartesian angle (measured counterclockwise from positive x-axis): North is 90°, so bearing 290° corresponds to angle: $$\theta_{AB} = 360^\circ - 70^\circ = 290^\circ$$ In Cartesian coordinates, angle from x-axis is: $$\alpha = 90^\circ + 70^\circ = 160^\circ$$ So vector AB has angle 160° from positive x-axis. Coordinates of B: $$B_x = 1600 \cos 160^\circ = 1600 \times (-0.9397) = -1503.5$$ $$B_y = 1600 \sin 160^\circ = 1600 \times 0.3420 = 547.2$$ 7. **Step 4: Use Law of Cosines to find angle at B:** $$\cos B = \frac{AB^2 + BC^2 - AC^2}{2 \times AB \times BC} = \frac{1600^2 + 600^2 - 1400^2}{2 \times 1600 \times 600}$$ Calculate numerator: $$1600^2 = 2,560,000$$ $$600^2 = 360,000$$ $$1400^2 = 1,960,000$$ $$\Rightarrow 2,560,000 + 360,000 - 1,960,000 = 960,000$$ Denominator: $$2 \times 1600 \times 600 = 1,920,000$$ So: $$\cos B = \frac{960,000}{1,920,000} = 0.5$$ Therefore: $$\angle B = \cos^{-1}(0.5) = 60^\circ$$ 8. **Step 5: Find coordinates of C:** We know: - Length BC = 600 ft - Angle at B = 60° - Vector AB angle = 160° Vector BC makes an angle of $160^\circ - 60^\circ = 100^\circ$ from positive x-axis (since angle at B is between AB and BC). Coordinates of C: $$C_x = B_x + 600 \cos 100^\circ = -1503.5 + 600 \times (-0.1736) = -1503.5 - 104.2 = -1607.7$$ $$C_y = B_y + 600 \sin 100^\circ = 547.2 + 600 \times 0.9848 = 547.2 + 590.9 = 1138.1$$ 9. **Step 6: Find bearing BC:** Vector BC components: $$\Delta x = C_x - B_x = -1607.7 + 1503.5 = -104.2$$ $$\Delta y = C_y - B_y = 1138.1 - 547.2 = 590.9$$ Calculate angle from north clockwise: Angle from x-axis: $$\theta = \tan^{-1}\left(\frac{590.9}{-104.2}\right) = \tan^{-1}(-5.67)$$ Since $\Delta x < 0$ and $\Delta y > 0$, vector is in second quadrant. $$\theta = 180^\circ - 80^\circ = 100^\circ$$ Bearing from north clockwise: $$\text{bearing} = 90^\circ - \theta = 90^\circ - 100^\circ = -10^\circ = 350^\circ$$ This means N10°E (10° east of north). 10. **Step 7: Find bearing CA:** Vector CA components: $$\Delta x = A_x - C_x = 0 - (-1607.7) = 1607.7$$ $$\Delta y = A_y - C_y = 0 - 1138.1 = -1138.1$$ Angle from x-axis: $$\theta = \tan^{-1}\left(\frac{-1138.1}{1607.7}\right) = \tan^{-1}(-0.708) = -35.3^\circ$$ Since $\Delta x > 0$ and $\Delta y < 0$, vector is in fourth quadrant. Bearing from north clockwise: $$\text{bearing} = 90^\circ - (-35.3^\circ) = 125.3^\circ$$ This is S55.3°E (since 125.3° is 35.3° east of south). **Final answers:** - Bearing BC = N10°E - Bearing CA = S55.3°E