Answer to Question #216564 in Geometry for Ajaja

Question

a ship leaves port at 1 pm traveling north at the speed of 30 miles per hour. At 3 pm the ship adjust its course 20 degrees eastward, how far was the ship from the port at 4pm. What law we will use to answer the problem

Answer 1

$$c=(3-1)\cdot30=60(mi)$$

$$a=(4-3)\cdot30=30(mi)$$

$$\beta=20\degree$$

The Law of Cosines

$$b^2=a^2+c^2-2ac\cos(180\degree-\beta)$$

$$b=\sqrt{30^2+60^2+2(30)(60)\cos20\degree}\approx88.786 (mi)$$