Airport's 80482

The plane flew from airport m on a course of 132° to airport n, then from n to p on a course of 235°. The distance between the airport's mn is 380 km, np 284 km. What will be the return course to m, and what is the distance between the airport's pm?

Correct answer:

φ = 13.8016 °
n = 420.12 km

Step-by-step explanation:

φ_{1} = 132^{\circ} φ_{2} = 235^{\circ} p = 380 km m = 284 km ∠ P N M = 180 - (φ_{2} - φ_{1}) = 180 - (235 - 132) = 77^{\circ} n^{2} = p^{2} + m^{2} - 2 \cdot p \cdot m \cdot cos P N M n = p^{2} + m^{2} - 2 \cdot p \cdot m \cdot cos P N M = p^{2} + m^{2} - 2 \cdot p \cdot m \cdot cos 77 ° = 38 0^{2} + 28 4^{2} - 2 \cdot 380 \cdot 284 \cdot cos 77 ° = 38 0^{2} + 28 4^{2} - 2 \cdot 380 \cdot 284 \cdot 0.224951 = 420.12208 km m^{2} = p^{2} + n^{2} - 2 \cdot p \cdot n \cdot cos M P N ∠ M P N = \frac{1 8 0 °}{π} \cdot arccos (\frac{p ^{2} + n ^{2} - m ^{2}}{2 \cdot p \cdot n}) = \frac{1 8 0 °}{π} \cdot arccos (\frac{3 8 0 ^{2} + 4 2 0 . 1 2 2 1 ^{2} - 2 8 4 ^{2}}{2 \cdot 3 8 0 \cdot 4 2 0 . 1 2 2 1}) ≐ 41.1984^{\circ} ∠ P M N = 180 - ∠ M P N - ∠ P N M = 180 - 41.1984 - 77 ≐ 61.8016^{\circ} φ = φ_{2} - 180 - ∠ M P N = 235 - 180 - 41.1984 = 13.801 6^{\circ} = 13 ° 4 8^{'} 6 "

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