Two ships leave a port from a point at the same time. One goes with a velocity of $3 \mathrm{~km} / \mathrm{h}$ along North-East maling an angle of $45^{\circ}$ with East direction and the other travels with a velocity of $4 \mathrm{~km} / \mathrm{h}$ along South-East maling an angle of $15^{\circ}$ with East direction. Then, the distance between the ships at the end of two hours is

• Correct answer
  A

  $2 \sqrt{13}$
• B

  $\sqrt{13}$
• C

  5
• D

  10

Solution

Distance travelled by ship $O A$ direction in 2 h is $6 \mathrm{~km}$ and $O B$ direction in $2 \mathrm{~h} 138 \mathrm{~km}$.
$\begin{aligned}<br />& A B^2=O A^2+O B^2-20 A O B \cos \angle A O B \\<br />& A B^2=(6)^2+(8)^2-2 \times 6 \times 8 \cos 60^{\circ} \\<br />& A B^2=36+64-2 \times 6 \times 8 \times \frac{1}{2}<br />\end{aligned}$

$\begin{gathered}<br />A B^2=100-48 \\<br />A B=\sqrt{100-48}=\sqrt{52} \\<br />A B=2 \sqrt{13}<br />\end{gathered}$

Hence, the correct answer is option: 'A'.