## Trigonometry (11th Edition) Clone

Let A be the ship's first position. Let B be the ship's final position. Let C be the position of the lighthouse. The points ABC form a triangle. The angle $A = 30^{\circ}$ The angle $B = 180^{\circ}-55^{\circ} = 125^{\circ}$ The angle $C = 180^{\circ}-30^{\circ}-125^{\circ} = 25^{\circ}$ Let $c = 2.0~mi$. Let $a$ be the distance from the ship's final position to the lighthouse. We can use the law of sines to find $a$: $\frac{a}{sin~A} = \frac{c}{sin~C}$ $a = \frac{c~sin~A}{sin~C}$ $a = \frac{(2.0~mi)~sin~30^{\circ}}{sin~25^{\circ}}$ $a = 2.4~mi$ The ship is 2.4 miles from the lighthouse.