1. **State the problem:** We are given two points with latitudes and longitudes: $$\lambda_1 = 00^\circ 06.0' N, \quad \lambda_1 = 119^\circ 05.0' E$$ $$\lambda_2 = 01^\circ 32.0' N, \quad \lambda_2 = 120^\circ 33.0' E$$ We want to find the distance $l$ and the angle $\lambda$ between these points using the formulas: $$\Phi = D l_0 \cos l_2$$ $$P = ?$$ $$\tan \lambda = \frac{P}{l}$$ where $D$ is the difference in latitude, $d lat$, and $l_0$ and $l_2$ are related to the longitudes. 2. **Convert coordinates to decimal degrees:** $$\lambda_1 = 0 + \frac{6.0}{60} = 0.1^\circ N$$ $$\lambda_2 = 1 + \frac{32.0}{60} = 1.5333^\circ N$$ $$l_1 = 119 + \frac{5.0}{60} = 119.0833^\circ E$$ $$l_2 = 120 + \frac{33.0}{60} = 120.55^\circ E$$ 3. **Calculate difference in latitude $D$:** $$D = \lambda_2 - \lambda_1 = 1.5333 - 0.1 = 1.4333^\circ$$ 4. **Calculate difference in longitude $d l$:** $$d l = l_2 - l_1 = 120.55 - 119.0833 = 1.4667^\circ$$ 5. **Calculate $l$ using the formula:** Assuming $l_0$ is the average latitude: $$l_0 = \frac{\lambda_1 + \lambda_2}{2} = \frac{0.1 + 1.5333}{2} = 0.8167^\circ$$ Convert $l_0$ to radians for cosine: $$l_0^{rad} = 0.8167 \times \frac{\pi}{180} = 0.01425$$ Calculate $l$: $$l = D \times l_0 \times \cos l_2$$ Since $l_2$ is longitude, convert to radians: $$l_2^{rad} = 120.55 \times \frac{\pi}{180} = 2.104$$ Calculate: $$l = 1.4333 \times 0.8167 \times \cos(2.104)$$ $$\cos(2.104) = -0.513$$ $$l = 1.4333 \times 0.8167 \times (-0.513) = -0.601$$ 6. **Calculate $P$:** Assuming $P = d lat / \cos l_0$: Convert $D$ to radians: $$D^{rad} = 1.4333 \times \frac{\pi}{180} = 0.0250$$ Calculate: $$P = \frac{D^{rad}}{\cos(l_0^{rad})} = \frac{0.0250}{\cos(0.01425)} = \frac{0.0250}{0.9999} = 0.0250$$ 7. **Calculate $\tan \lambda$ and $\lambda$:** $$\tan \lambda = \frac{P}{l} = \frac{0.0250}{-0.601} = -0.0416$$ Calculate $\lambda$: $$\lambda = \arctan(-0.0416) = -2.38^\circ$$ **Final answers:** $$l = -0.601$$ $$P = 0.0250$$ $$\lambda = -2.38^\circ$$ This means the angle $\lambda$ is approximately $-2.38^\circ$, and the distance $l$ is about $-0.601$ (units depend on context).