Question

Two ships leave a port at the same time. One of them moves in the direction of E50$^\circ$N with a speed of 8 kmph and the other moves in the direction of S20$^\circ$E with a speed of 12 kmph. Then the distance between the ships at the end of 2 hours is (in km)

• A. $8\sqrt{7}$
• B. 34
• C. $8\sqrt{19}$
• D. 32

Hint:

In problems involving bearings and distances, draw a diagram with a coordinate system (N-E-S-W). Convert the bearings to standard angles and use the Law of Cosines to find the distance between the final points.

Answer 1

The Correct Option is C

Let the port be the origin O. Let the positions of the two ships after 2 hours be A and B.
Distance of Ship 1 from the port: $OA = \text{speed} \times \text{time} = 8 \text{ kmph} \times 2 \text{ h} = 16 \text{ km}$.
Distance of Ship 2 from the port: $OB = \text{speed} \times \text{time} = 12 \text{ kmph} \times 2 \text{ h} = 24 \text{ km}$.
Now, we need to find the angle between their paths, $\angle AOB$.
The direction of Ship 1 is E50$^\circ$N, which means 50$^\circ$ North of the East direction. The angle with the positive x-axis (East) is $50^\circ$.
The direction of Ship 2 is S20$^\circ$E, which means 20$^\circ$ East of the South direction. The angle measured clockwise from South is 20$^\circ$. The South direction is at $-90^\circ$. So, the angle of Ship 2 is $-90^\circ + 20^\circ = -70^\circ$ from the positive x-axis.
The total angle between their paths is the difference between their angles: $\angle AOB = 50^\circ - (-70^\circ) = 120^\circ$.
We now have a triangle OAB with sides $OA=16$, $OB=24$, and the included angle $\angle AOB = 120^\circ$. We need to find the length of the third side, AB.
Using the Law of Cosines:
$AB^2 = OA^2 + OB^2 - 2(OA)(OB)\cos(\angle AOB)$.
$AB^2 = 16^2 + 24^2 - 2(16)(24)\cos(120^\circ)$.
$AB^2 = 256 + 576 - 2(16)(24)(-\frac{1}{2})$.
$AB^2 = 832 + (16)(24) = 832 + 384 = 1216$.
The distance is $AB = \sqrt{1216}$.
To simplify the radical: $1216 = 64 \times 19$.
$AB = \sqrt{64 \times 19} = 8\sqrt{19}$ km.