Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2632 Upvotes
Verified Answer
The correct answer is:
$2 \sqrt{13}$
Distance travelled by ship $O A$ direction in 2 h is $6 \mathrm{~km}$ and $O B$ direction in $2 \mathrm{~h} 138 \mathrm{~km}$.
$\begin{aligned}
& A B^2=O A^2+O B^2-20 A O B \cos \angle A O B \\
& A B^2=(6)^2+(8)^2-2 \times 6 \times 8 \cos 60^{\circ} \\
& A B^2=36+64-2 \times 6 \times 8 \times \frac{1}{2}
\end{aligned}$
$\begin{gathered}
A B^2=100-48 \\
A B=\sqrt{100-48}=\sqrt{52} \\
A B=2 \sqrt{13}
\end{gathered}$
$\begin{aligned}
& A B^2=O A^2+O B^2-20 A O B \cos \angle A O B \\
& A B^2=(6)^2+(8)^2-2 \times 6 \times 8 \cos 60^{\circ} \\
& A B^2=36+64-2 \times 6 \times 8 \times \frac{1}{2}
\end{aligned}$
$\begin{gathered}
A B^2=100-48 \\
A B=\sqrt{100-48}=\sqrt{52} \\
A B=2 \sqrt{13}
\end{gathered}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.