## Trigonometry (11th Edition) Clone

Let A be the starting point. Let B be the point where the ship turns. Let C be the final point. The pints A, B, and C form a triangle. The length of AB is 18.5 km, The length of BC is 47.8 km. We can find the angle of the triangle at the point B. Angle $B = (90^{\circ}-47^{\circ})+9^{\circ} = 52^{\circ}$ We can use the law of cosines to find the length $AC$, which is the distance of the ship from point A: $AC^2 = AB^2+BC^2-2(AB)(BC)~cos~B$ $AC = \sqrt{AB^2+BC^2-2(AB)(BC)~cos~B}$ $AC = \sqrt{(18.5~km)^2+(47.8~km)^2-(2)(18.5~km)(47.8~km)~cos~52^{\circ}}$ $AC = \sqrt{1538.23~km^2}$ $AC = 39.2~km$ The distance of the ship from point A is 39.2 km