A coastal monitoring station at point O tracks two vessels on horizontal water. Vessel A i… — JAMB UTME

A coastal monitoring station at point $O$ tracks two vessels on horizontal water. Vessel $A$ is located $15\text{ km}$ from $O$ on a bearing of $070^\circ$ , while Vessel $B$ is located $20\text{ km}$ from $O$ on a bearing of $160^\circ$ . What is the direct distance between Vessel $A$ and Vessel $B$ in kilometers?

Answer: 25 km

Answer

The direct distance between Vessel A and Vessel B is 25 km.

The difference between the two bearings (

160^\circ - 070^\circ = 90^\circ

) establishes that triangle

AOB

is a right-angled triangle at station

O

. Applying Pythagoras' theorem yields

AB = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25\text{ km}

Step-by-Step Solution

Find the angle between the lines of sight to the two vessels

\angle AOB = 160^\circ - 070^\circ = 90^\circ

Subtracting the smaller bearing angle from the larger bearing angle from the same origin point gives the included angle.

Set up the equation for distance AB using Pythagoras' theorem

AB^2 = 15^2 + 20^2 = 225 + 400 = 625

Since the included angle is 90 degrees, the three points form a right-angled triangle where AB is the hypotenuse.

Calculate the principal square root of 625

AB = \sqrt{625} = 25\text{ km}

Taking the square root converts the squared distance into the direct linear distance between the vessels.

Key Concept

Calculating the distance between two points using bearings and right-angled triangle properties (Pythagoras' theorem).

Estimated Time:1m 30s

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