A boat radioed a distress call to a Coast Guard station. At the time of the call, a vector A from the station to the boat had a magnitude of 45.0 km and was directed 15.0° east of north. A vector from the station to the point where the boat yxas later found is =30.0 km, 15.0° north of east.
-How far did the boat travel from the point where the distress call was made to the point where the boat was found? In other words, what is the magnitude of vector ?
A)65.3 km
B)39.7 km
C)26.5 km
D)54.0 km
E)42.5 km

Question

A boat radioed a distress call to a Coast Guard station. At the time of the call, a vector A from the station to the boat had a magnitude of 45.0 km and was directed 15.0° east of north. A vector from the station to the point where the boat yxas later found is =30.0 km, 15.0° north of east.
-How far did the boat travel from the point where the distress call was made to the point where the boat was found? In other words, what is the magnitude of vector  ?
A)65.3 km 
B)39.7 km 
C)26.5 km 
D)54.0 km 
E)42.5 km

Quizplus · Accepted Answer

We can use the law of cosines to find the magnitude of the vector connecting the distress call location to the location where the boat was found. Let's call this vector C.

cos(angle between A and C) = (A dot C)/(magnitude of A * magnitude of C)

cos(105°) = (45.0 km * 30.0 km * cos(15°) + 45.0 km * C)/sqrt((45.0 km)^2 + (30.0 km)^2 + 2(45.0 km)(30.0 km)cos(15°))

Solving for C gives:

C = 39.7 km

So the boat traveled 39.7 km from the distress call location to the location where it was found. The answer is therefore (B).

A Boat Radioed a Distress Call to a Coast Guard