The magnitude of a force is 420 lb and the direction is 128

The magnitude of a force is 420 lb and the direction is 128 degree with the horizontal. Find the horizontal and vertical components of the force, rounding to the nearest tenth. Given v = , find | v| and the angle, rounding to the nearest degree. If v = and w : . find the angle between d and w. 10. A boot travels 128 miles on a course of bearing N68 degree W and then changes its course to travel 52 miles at $22 degree E. Draw the vectors and use Law of Cosines to find the distance from the final point to the starting point. Find the bearing from the final point to the starting point.

Solution

post 10th question separately

Force magnithude=420 lb

vertical component = F cos128

= -259 lb

horizontal component = F sin128

=331 lb

8). V=<-6,14>

then |v| = sqrt(6^2 +14^2)

=sqrt(36 +196)

=sqrt(232)

angle tan(x) = 14/-6

angle = 113 degree

9).

v=<2,10> w=<-4,2>

v.w =|v| .|w| cos(x)

<2,10>.<-4,2> = sqrt(4+100) sqrt(16+4) cos(x)

-8 +20 = sqrt(104) .sqrt(20) cos(x)

12 = sqrt(104) .sqrt(20) cos(x)

cos(x) =0.26

x=74.7 degree

angle between v and w is 75 degree (approximately)