Two ships A and B are sailing straight away from a fixed point $\mathrm{O}$ along routes such that $\angle \mathrm{AOB}$ is always $120^{\circ}$. At a certain instance, $\mathrm{OA}=8 \mathrm{~km}, \mathrm{OB}=6 \mathrm{~km}$ and the ship $\mathrm{A}$ is sailing at the rate of $20 \mathrm{~km} /…

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Two ships A and B are sailing straight away from a fixed point $\mathrm{O}$ along routes such that $\angle \mathrm{AOB}$ is always $120^{\circ}$. At a certain instance, $\mathrm{OA}=8 \mathrm{~km}, \mathrm{OB}=6 \mathrm{~km}$ and the ship $\mathrm{A}$ is sailing at the rate of $20 \mathrm{~km} / \mathrm{hr}$ while the ship B sailing at the rate of $30 \mathrm{~km} / \mathrm{hr}$. Then the distance between $\mathrm{A}$ and $\mathrm{B}$ is changing at the rate (in $\mathrm{km} / \mathrm{hr}$ ):

Aicharya · Accepted Answer

Let $\mathrm{OA}=x \mathrm{~km}, \mathrm{OB}=y \mathrm{~km}, \mathrm{AB}=\mathrm{R}$ $(\mathrm{AB})^2=(\mathrm{OA})^2+(\mathrm{OB})^2-2(\mathrm{OA})(\mathrm{OB}) \cos$ $120^{\circ}$ $ \mathrm{R}^2=x^2+y^2-2 x y\left(-\frac{1}{2}\right)=x^2+y^2+x y $ $\mathrm{R}$ at $x=6 \mathrm{~km}$, and $y=8 \mathrm{~km}$ $ \mathrm{R}=\sqrt{6^2+8^2+6 \times 8}=2 \sqrt{37} $ Differentiating equation (1) with respect to $t$ $ \begin{gathered} 2 R \frac{d R}{d t}=2 \frac{d x}{d t}+2 y \frac{d y}{d t}+\left(x \frac{d y}{d t}+y \frac{d x}{d t}\right) \\ =\frac{1}{2 R}[2 \times 8 \times 20+2 \times 6 \times 30+(8 \times 30+6 \times 20)] \\ \frac{d R}{d t}=\frac{1}{2 \times 2 \sqrt{37}}[1040]=\frac{260}{\sqrt{37}} \end{gathered} $

Two ships A and B are sailing straight away from a fixed point $\mathrm{O}$ along routes such that $\angle…

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