SOLUTION: Two boats left the harbour at the same time. One travelled at 10km/h on a bearing pf N47'E. The other travelled at 8km/h on a bearing of N79'E. How far apart were the boats after 4

Algebra ->  Triangles -> SOLUTION: Two boats left the harbour at the same time. One travelled at 10km/h on a bearing pf N47'E. The other travelled at 8km/h on a bearing of N79'E. How far apart were the boats after 4      Log On


   

Question 1107840: Two boats left the harbour at the same time. One travelled at 10km/h on a bearing pf N47'E. The other travelled at 8km/h on a bearing of N79'E. How far apart were the boats after 45 min? Round your answer to the nearest tenth of a kilometre.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Two boats left the harbour at the same time.
One travelled at 10km/h on a bearing pf N47'E.
The other travelled at 8km/h on a bearing of N79'E.
How far apart were the boats after 45 min?
Round your answer to the nearest tenth of a kilometre.
:
the angle between the paths of the two boats: 79 - 47 = 32 degrees
find how far each boat has traveled in 45 min (3%2F4 hr)
Boat l:3%2F4 * 10 = 7.5 km
Boat 2: 3%2F4 * 8 = 6 km
Use the law of cosines a^2 = b^2 + c^2 - 2(b*c)*cos(A) where
a = distance between the boats after 45 min
b = 7.5 km
c = 6 km
A = 32 degrees
a^2 = 7.5^2 + 6^2 - 2(7.5*6)*cos(32)
a^2 = 56.25 + 36 - 2(45)*.848
Do the math, I got:
a = sqrt%2815.92567%29
a = 4.0 km