## Trigonometry (11th Edition) Clone

Let A be the point where the mountain is at a bearing of $24.1^{\circ}$ Let B be the point where the mountain is at a bearing of $32.7^{\circ}$ Let point C be the location of the mountain. The points ABC form a triangle. The angle $A = 24.1^{\circ}$ The angle $B = 180^{\circ}-32.7^{\circ} = 147.3^{\circ}$ We can use the law of sines to find the distance $BC$, which is the distance from the mountain when the second bearing is taken: $\frac{BC}{sin~24.1^{\circ}} = \frac{7.92}{sin~147.3^{\circ}}$ $BC = \frac{7.92~sin~24.1^{\circ}}{sin~147.3^{\circ}}$ $BC = 5.99~km$ The distance between the airplane and the mountain is 5.99 km when the second bearing is taken.