Two ships, P and Q, leave a port O at the same time. Ship P sails on a bearing of 040^\cir… — JAMB UTME

Two ships, $P$ and $Q$ , leave a port $O$ at the same time. Ship $P$ sails on a bearing of $040^\circ$ at a constant speed of $25\text{ km/h}$ , while Ship $Q$ sails on a bearing of $100^\circ$ at a constant speed of $40\text{ km/h}$ . What is the distance in kilometers between the two ships after $2$ hours?

Answer: 70 km

Answer

The distance between the two ships after 2 hours is 70 km.

The distance traveled by Ship P in 2 hours is

50\text{ km}

and by Ship Q is

80\text{ km}

. The angle between their paths is

100^\circ - 040^\circ = 60^\circ

. Applying the Cosine Rule yields

PQ^2 = 50^2 + 80^2 - 2(50)(80)\cos(60^\circ) = 2500 + 6400 - 4000 = 4900

, giving a distance of

\sqrt{4900} = 70\text{ km}

Step-by-Step Solution

Calculate the distances traveled by Ship P and Ship Q after 2 hours

OP = 50\text{ km}

and

OQ = 80\text{ km}

Distance equals speed multiplied by time.

Find the angle between the lines of travel from port O

\angle POQ = 100^\circ - 40^\circ = 60^\circ

The angle between two bearings from a common origin is the difference between their bearing angles.

Use the Cosine Rule to calculate the side length PQ

PQ^2 = 50^2 + 80^2 - 2(50)(80)\cos(60^\circ) = 4900

The Cosine Rule

c^2 = a^2 + b^2 - 2ab\cos C

calculates the unknown opposite side given two side lengths and their included angle.

Take the square root to find PQ

PQ = 70\text{ km}

Taking the principal square root gives the final linear distance.

Key Concept

Applying the Cosine Rule to solve bearing and distance non-right triangle problems

Estimated Time:2m 0s

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