At midnight, a car leaves an intersection heading due east at a constant speed of 62 mph. At the same time, a bicycle leaves the intersection on a second road at a constant speed of 20 mph.

Question

 At midnight, a car leaves an intersection heading due east at a constant speed of 62 mph. At the same time, a bicycle leaves the intersection on a second road at a constant speed of 20 mph. At 1:00 AM it is observed that the distance between the bicycle and car is increasing at the rate of 49 mph. Find the angle of intersection, to the nearest degree, between the two roads. Hint: Use the law of cosines

Answer 1

law of cosines is:

c² = a² + b² - 2ab*cos(θ)

the main point of this question is highlighting a few things:

1. velocity is a vector, and we can use trig style functions to solve it, like the law of cosines
2. the θ we're solving for is the angle between vectors a and b.
3. plug in the correct values we're given

since we want the angle between the car and bicycle, c is the *net*:

49^2 = 62^2 + 20^2 - 2*20*62*cos(θ)

then we can get cos(θ) by itself.

(62^2 + 20^2 - 49^2) / (2*20*62 = cos(θ) = 0.74

don't forget to convert to degrees:

θ = arccos(0.74) * 180 / 2π = 21^ο

fun note:

since the car is going east, this result says that the bike is going 21^ο with respect to east. This could mean that the bike is either north-east or south-east.