Altitude angles

Cities A, B, and C lie in one elevation plane. C is 50 km east of B, and B is north of A. C is deviated by 50° from A. The plane flies around places A, B, and C at the same altitude. When the aircraft is flying around B, its altitude angle to A is 12°. Find the altitude angle to A as the aircraft flies around city C.

Correct answer:

C =  7.7801 °

Step-by-step explanation:

BC=50 km α=50  A=12   β=90  sin α = BC/AC  AC=BC/sinα=BC/sin50° =50/sin50° =50/0.766044=65.27036 km AB=AC2BC2=65.2704250241.955 km  tan A = h : AB  h=AB tanA=AB tan12° =41.955 tan12° =41.955 0.212557=8.91781 km  tan C = h : AC  C=π180°arctan(h/AC)=π180°arctan(8.9178/65.2704)=7.7801=7°4648"

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