Question

could I have the triangle drawn for this problem. Also show which side is A, b, c and why.

A friend walks away from you a distance of 540m, and then turns (as if on a dime) an unknown angle, and walks in an additional 245 m in the new direction. you use a laser range finder to find out that his final distance from you is 432m.whats the angle between his initial departure direction and the direction to his final location.(enter only magnitude)

Answer: 26.3 degree, 129 degree

Answer 1

Let his initial direction was along x axis and he turned by angle theta

By vector addition,

Resultant = sqrt(a^2+b^2+2ab cos theta), here a is first vector, b is second vector, c is resultant.

432 = sqrt(540^2+245^2+2*540*245* cos theta)

432^2 = 540^2+245^2+2*540*245* cos theta

cos theta = -0.6236 , theta = 129 degree

angle between his initial departure and final location is given by cosine rule,

b^2 = a^2+c^2 +2ac cos B where B is angle opposite b, i.e. angle between his initial departure and final location direction

245^2=540^2+432^2+2*540*432* cos B

cos B = 0.896

B = arccos 0.896 = 26.3 degree answer