Two sailboats leave a harbor in the bahamas at the same time. the first sails at 23 mph in a direction 330°. the second sails at 34 mph in a direction 190°. assuming that both boats maintain speed and heading, after 2 hours, how far apart are the boats?

First, let us calculate the total distance that each have taken after 2 hours.

Let’s say that:

A = sailboat which sails at 23 mph in a direction 330°

B = sailboat which sails at 34 mph in a direction 190°

Calculating for distances:

dA = 23 mph (2 hours) = 46 miles

dB = 34 mph (2 hours) = 68 miles

Imagining a Cartesian coordinate, the angle θ between the two sailboats is simply the difference:

θ = 330° - 190°

θ = 140°

We know that from the law of cosines:

c^2 = a^2 + b^2 – 2 a*b*cos θ

Therefore the distance between the two after 2 hours, C, is:

C^2 = 46^2 + 68^2 – 2 (46) (68) cos(140)

C = 107.39 miles

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..

the components of a vector are the parts of the vectors that are to each other.

b perpendicular