Answer to Question #125187 in Physics for qwerty

Question

A sailor in a small boat encounters shifting winds. She sails 8 km south, then 15 km 30 E of N, and then 12 km 25 N of W. Use any method to determine the magnitude of her resultant displacement in kilometers

Answer 1

The resultant displacement is shown on the figure (OC).

Let's denote the origin as O. Then, using the law of cosines, from the triangle AOB find:

$$OB^2 = OA^2 + AB^2 - 2OA\cdot AB\cdot \cos(30\degree) = 8^2 + 15^2 - 2\cdot 8\cdot 15\cdot \dfrac{\sqrt{2}}{2} \approx119.2\\ OB \approx 10.9 km$$

Using the law of sines, from the triangle AOB find:

$$\angle OBA = \arcsin{(\dfrac{8km\cdot \sin 30\degree}{10.9 km})} \approx 21.5\degree$$

From the right triangle ADB, the angle DBA is equal to 180-90-30 = 60 degrees. Thus, the angle CBO will be:

$$\angle CBO = \angle DBA+25\degree-\angle OBA = 60\degree+25\degree-21.5\degree = 63.5\degree$$

Then, using the law of cosines, from the triangle OCB find:

$$OC^2 = OB^2 + CB^2 - 2OB\cdot CB\cdot \cos(63.5\degree) =\\ = 10.9^2 + 12^2 - 2\cdot 10.9\cdot 12\cdot \cos(63.5\degree)\approx116.9\\ OC \approx 10.8 km$$

Answer. The resultant displacement is 18.8 km.