9514 1404 393
Answer:
108.8 km/h at 84.3°
Step-by-step explanation:
The law of cosines can be used to find the resultant ground speed. In the attached diagram, the length of interest is OR. It will be found as ...
OR² = OP² +PR² -2·OP·PR·cos(P)
OR² = 130² +26² -2×130×26×cos(32°) ≈ 11843.19
OR = √11843.19 ≈ 108.8
__
The angle POR can be found from the law of sines.
sin(POR)/PR = sin(OPR)/OR
sin(POR) = PR/OR×sin(32°) ≈ 0.12660
∠POR ≈ arcsin(0.12660) ≈ 7.27°
Then the bearing of the ground track of the airplane is 77° +7.27° = 84.27°.
The airplane is traveling at about 108.8 km/h on a bearing of 84.3°.
