Help! Stuck on Trig Question.

[StellaS]

I'm a high school student who's been stuck on this question for days.

A plane leaves an aircraft carrier and flies due north at 500km/h. The aircraft carrier proceeds at 30 degrees west of south at 35km/h. If the plane has enough fuel for 4 hours of flying, what is the maximum distance north it can fly so that the fuel remaining will allow a safe return to the aircraft carrier?

Your help is greatly appreciated.

 

[galactus]

The total distance the plane can fly is 500(4)=2000 km.

Use the law of cosines:

\(\displaystyle \L\\(2000-500x)^{2}=(35x)^{2}+(500x)^{2}-2(35x)(500x)cos(150)\)

\(\displaystyle \L\\250000x^{2}-2000000x+4000000=(17500\sqrt{3}+251225)x^{2}\)

\(\displaystyle \L\\\frac{-7(100\sqrt{3}+7)}{40000}x^{2}-2x+4=0\)

solve for x.

Multiply it by 500 to see how far the planr can fly north before it must turn

back toward the carrier.

 

[StellaS]

Thank you so much for your help!

 

[Denis]

Another way: make the triangle "manageable"; plane travels from A to B, back to C where carrier is (carrier travels A to C):

        A


C                                                              B

AC = b = 140
BC = a
AB = c = 2000 - a
Angle A = 150 degrees

a^2 = b^2 + c^2 - 2bc COS(A)
a^2 = 140^2 + (2000 - a)^2 - 2(140)(2000 - a) COS(150)

Carry on...the a^2 will cancel out, so you can then solve directly.