## Trigonometry (11th Edition) Clone

The distance between the ship and the rock when the ship is at its final position is $~~10.8~miles$
Let $A$ be the initial position of the ship. Let $B$ be the final position of the ship. Let $C$ be the position of the rock. Then the points $A,$ $B,$ and $C$ form a triangle. The angle at point $A$ is $90^{\circ} - 45^{\circ}20' = 44^{\circ}40'$ The angle at point $B$ is $308^{\circ}40'-270^{\circ} = 38^{\circ}40'$ The angle at point $C$ is $180^{\circ}-44^{\circ}40'-38^{\circ}40' = 96^{\circ}40'$ We can find $b$, the side of the triangle subtended by angle $B$: $\frac{b}{sin~B} = \frac{c}{sin~C}$ $b = \frac{c~sin~B}{sin~C}$ $b = \frac{(15.2)~sin~38^{\circ}40'}{sin~96^{\circ}40'}$ $b = 9.56~mi$ We can use the law of cosines to find $a$, the distance between the ship and the rock when the ship is at its final position: $a^2 = b^2+c^2-2bc~cos~A$ $a^2 = (9.56)^2+(15.2)^2-2(9.56)(15.2)~cos~44^{\circ}40'$ $a^2 = 115.739$ $a = 10.8~miles$ The distance between the ship and the rock when the ship is at its final position is $~~10.8~miles$