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luo's question at Yahoo! Answers regarding an application of the Law of Cosines
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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 \(\displaystyle (180-30)^{\circ}=150^{\circ}\). 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:
\(\displaystyle (1600-D)^2=120^2+D^2-2\cdot120\cdot D\cdot\cos\left(150^{\circ} \right)\)
Expanding on the left, and using \(\displaystyle \cos\left(150^{\circ} \right)=-\frac{\sqrt{3}}{2}\), we obtain:
\(\displaystyle 1600^2-3200+D^2=120^2+D^2+120\sqrt{3}D\)
Solving for $D$, we obtain:
\(\displaystyle D=\frac{1600^2-120^2}{3200+120\sqrt{3}}=\frac{63640}{6373}\left(80-3\sqrt{3} \right)\approx747\)
Hence, the aircraft can fly about 747 kilometers due south before having to retunr to the carrier.
We know the angle subtending the carrier's position vector and the position vector of the aircraft's first leg of its trip is \(\displaystyle (180-30)^{\circ}=150^{\circ}\). 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:
\(\displaystyle (1600-D)^2=120^2+D^2-2\cdot120\cdot D\cdot\cos\left(150^{\circ} \right)\)
Expanding on the left, and using \(\displaystyle \cos\left(150^{\circ} \right)=-\frac{\sqrt{3}}{2}\), we obtain:
\(\displaystyle 1600^2-3200+D^2=120^2+D^2+120\sqrt{3}D\)
Solving for $D$, we obtain:
\(\displaystyle D=\frac{1600^2-120^2}{3200+120\sqrt{3}}=\frac{63640}{6373}\left(80-3\sqrt{3} \right)\approx747\)
Hence, the aircraft can fly about 747 kilometers due south before having to retunr to the carrier.