Look at ' Anything else you'd like to explain about your question' to view the entire question.A plane leaves an aircraft carrier and flies south...

A plane leaves an aircraft carrier and flies
south at 800km/hr while the carrier proceeds in the direction 60 degrees west of north
at 40km/hr. If the plane has enough fuel to fly 6 hrs, how far south can it fly before
returning to the carrier? Find also the distance travelled by the plane during this
flight.

Let a represent the distance the plane
flew due south, b the distance the carrier travelled, and c the distance from the
furthest point south back to the carrier.

The angle between
sides a and b of the triangle is the supplement of 60 degrees, or 120
degrees.

We use the Law of Cosines: `c^2=a^2+b^2-2abcosC`
where angle C has measure 120 degrees.

We assume the plane
uses all of the available time, thus the total flight time is 6 hours. Let t be the time
the plane flies due south. Then:

(1) b=(40
km/hr)(6hr)=320km

(2) a=(800 km/hr)(t hr)=800t
km

(3) c=(800 km/hr)((6-t) hr)=(4800-800t)km *time on
return trip is 6hr minus the time flying due south *

Then
`(4800-800t)^2=(800t)^2+320^2-2(800t)(320)(cos120)`

`4800^2-2(4800)(800t)+(800t)^2=(800t)^2+320^2+(800t)(320)`

`4800^2-320^2=(800t)(320)+2(4800)(800t)`

`22937600=7936000t`

`2.89~~t`
. Thus the plane flew due south for about 2.89 hours, and flew for about 3.11 hours
returning to the carrier.

The distances are
`a~~2312km,b~~320km,c~~2488km` . The total distance flown is 4800km, which agrees with a
6hr flight at 800km/hr.