MHB Luo's question at Yahoo Answers regarding an application of the Law of Cosines

[MarkFL]

Here is the question:

A reconnaissance aircraft leaves aircraft carrier and flies south at a speed of 400kph. During this time the?

Carrier proceeds in direction of N30 degree W at speed of 30kph. If the pilot has enough fuel to fly for 4 hours, how far south can he fly before he has return to the ship.

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello luo,

We know the angle subtending the carrier's position vector and the position vector of the aircraft's first leg of its trip is $$(180-30)^{\circ}=150^{\circ}$$. We also know the magnitude of the carrier's position vector during the 4 hours is $120\text{ km}$. If we let $D$ represent the distance the aircraft may fly south before having to turn back to the carrier, and we know the total distance it can fly in the 4 hours is $1600\text{ km}$, then we knoe the magnitude of the second leg of the aircraft's journey is $(1600-D)\text{ km}$.

Using the Law of Cosines, we may then state:

$$(1600-D)^2=120^2+D^2-2\cdot120\cdot D\cdot\cos\left(150^{\circ} \right)$$

Expanding on the left, and using $$\cos\left(150^{\circ} \right)=-\frac{\sqrt{3}}{2}$$, we obtain:

$$1600^2-3200+D^2=120^2+D^2+120\sqrt{3}D$$

Solving for $D$, we obtain:

$$D=\frac{1600^2-120^2}{3200+120\sqrt{3}}=\frac{63640}{6373}\left(80-3\sqrt{3} \right)\approx747$$

Hence, the aircraft can fly about 747 kilometers due south before having to retunr to the carrier.