Question

Two bicyclists left at 11:30 AM for a ride. Rider A traveled at a constant rate of 12 mph at a
bearing of N 47° E, stopping at 1:30 PM. Rider B traveled until 3:00 PM, biking at a constant rate
of 9 mph at a bearing of S 62° E. Determine the distance between the two bicyclists.

Answer 1

Answer:

32.8 miles

Step-by-step explanation:

The total time travelled by rider A is 2 hours. If his speed was 12 mph, the distance travelled is 2 * 12 = 24 miles.

The total time travelled by rider A is 3 hours and 30 minutes (so 3.5 hours). If his speed was 9 mph, the distance travelled is 3.5 * 9 = 31.5 miles.

The angle that they make, as rider A was in the direction N 47 E and the rider B was in the direction S 62 E, is equal to 180 - 47 - 62 = 71°

With these values (two distances and a angle between then, we can use law of cosines to find the distance between the bicyclists:

c^2 = a^2 + b^2 - 2ab*cos(angle)

c^2 = 24^2 + 31.5^2 - 2*24*31.5*cos(71)

c^2 = 1075.99

c = 32.8 miles

Answer 2

The distance between the two bicyclist is 32.8 miles

• The calculation is as follows:

The total time travelled by rider A is 2 hours.

In the case when his speed was 12 mph,

So, the distance travelled is 2 (12) = 24 miles.

Now

The total time travelled by rider A is 3 hours and 30 minutes i.e. 3.5 hours).

In the case when his speed was 9 mph, the distance travelled is 3.5 (9) = 31.5 miles.

Now the direction is

= 180 - 47 - 62

= 71°

Now the distance is

$c^2 = a^2 + b^2 - 2ab\times cos(angle)\\\\c^2 = 24^2 + 31.5^2 - 2\times 24\times 31.5\times cos(71)\\\\c^2 = 1075.99$

c = 32.8 miles

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