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:

$$\begin{gathered} BC=50\text{km} \\ \alpha =50{}^{\circ } \\ A=12{}^{\circ } \\ \beta =90{}^{\circ } \\ \sin \alpha =BC/AC \\ AC=BC/\sin \alpha =BC/\sin 50°=50/\sin 50°=50/0.766044=65.27036\text{km} \\ AB=\sqrt{A{C}^2-B{C}^2}=\sqrt{65.2704^2-50^2}\doteq 41.955\text{km} \\ \tan A=h:AB \\ h=AB\cdot \tan A=AB\cdot \tan 12°=41.955\cdot \tan 12°=41.955\cdot 0.212557=8.91781\text{km} \\ \tan C=h:AC \\ C=\frac{180°}{\pi }\cdot \arctan (h/AC)=\frac{180°}{\pi }\cdot \arctan (8.9178/65.2704)=7.7801^{\circ }=7°46^{\prime }48" \end{gathered}$$

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