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

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The discussion centers on a problem involving a reconnaissance aircraft and its carrier, requiring the application of the Law of Cosines to determine how far south the aircraft can fly before needing to return. The carrier moves at a speed of 30 kph in a direction of N30°W, while the aircraft flies south at 400 kph for a maximum of 4 hours. The angle between the carrier's and aircraft's position vectors is calculated to be 150°, and using the Law of Cosines, the distance the aircraft can travel south is derived. The final calculation shows that the aircraft can fly approximately 747 kilometers south before needing to return to the carrier. This solution effectively applies trigonometric principles to solve the navigational problem.
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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.
 
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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.
 
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