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:

$$\begin{gathered} {\phi }_1=132{}^{\circ } \\ {\phi }_2=235{}^{\circ } \\ p=380\text{km} \\ m=284\text{km} \\ \angle PNM=180-({\phi }_2-{\phi }_1)=180-(235-132)=77{}^{\circ } \\ {n}^2=p^2+m^2-2\cdot p\cdot m\cdot \cos PNM \\ n=\sqrt{p^2+m^2-2\cdot p\cdot m\cdot \cos PNM}=\sqrt{p^2+m^2-2\cdot p\cdot m\cdot \cos 77°}=\sqrt{380^2+284^2-2\cdot 380\cdot 284\cdot \cos 77°}=\sqrt{380^2+284^2-2\cdot 380\cdot 284\cdot 0.224951}=420.12208\text{km} \\ {m}^2=p^2+n^2-2\cdot p\cdot n\cdot \cos MPN \\ \angle MPN=\frac{180°}{\pi }\cdot \arccos (\frac{p^2+n^2-m^2}{2\cdot p\cdot n})=\frac{180°}{\pi }\cdot \arccos (\frac{380^2+420.1221^2-284^2}{2\cdot 380\cdot 420.1221})\doteq 41.1984{}^{\circ } \\ \angle PMN=180-\angle MPN-\angle PNM=180-41.1984-77\doteq 61.8016{}^{\circ } \\ \phi ={\phi }_2-180-\angle MPN=235-180-41.1984=13.8016^{\circ }=13°48^{\prime }6" \end{gathered}$$

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