Question

Two airplanes leave an airport at the same time. The first flies $150\frac{km}{h}$ in a direction of $320°$. The second flies $200\frac{km}{h}$ in a direction of $200°$. After $3hr$ , how far apart are the planes?

Answer 1

Distance between the planes:

Difference of angles between the motion of the planes is given by:

$$\begin{gathered} \theta =320-200 \\ =120° \end{gathered}$$

Let $a$ be the distance travelled by the first fly,

Then $dis\tan ce=speed\times time$

$$\begin{gathered} \Rightarrow a=150\times 3 \\ \Rightarrow a=450km \end{gathered}$$

Let b be the distance travelled by the second fly,

$$\begin{gathered} \Rightarrow b=200\times 3 \\ \Rightarrow b=600km \end{gathered}$$

Now, the distance between the planes after $3$ hours will be equal to the side of a triangle opposite to the angle $\theta =120°$ included by the sides $a=450\&b=600$.

Now, using Cosine rule in the concerned triangle, the distance between the plane after $3$ hours is given as

$$\begin{gathered} =\sqrt{a^2+b^2-2ab\cos \theta } \\ =\sqrt{{450}^2+{600}^2-2\cdot 450\cdot 600\cos 120°} \\ =\sqrt{562500-540000\left(-\frac{1}{2}\right)} \\ =150\sqrt{13} \\ =540.832km \end{gathered}$$

Hence,$540.832km$ is the distance between the flies after $3$ hrs.