A cargo ship departs from port M and sails 24\text{ km} on a bearing of 050^\circ to reach… — JAMB UTME

A cargo ship departs from port $M$ and sails $24\text{ km}$ on a bearing of $050^\circ$ to reach point $N$ . From point $N$ , the ship changes course and sails $10\text{ km}$ on a bearing of $140^\circ$ to reach point $P$ . What is the direct distance, in kilometers, from port $M$ to point $P$ ?

Answer: 26 km

Answer

The direct distance from port M to point P is 26 km.

The back bearing from N to M is 230°, and the bearing from N to P is 140°. The interior angle at N is 230° - 140° = 90°. Using the Pythagorean theorem for the right triangle formed by M, N, and P, the direct distance is √(24² + 10²) = √676 = 26 km.

Step-by-Step Solution

Calculate the interior angle

\angle MNP

at point

N

\angle MNP = (050^\circ + 180^\circ) - 140^\circ = 230^\circ - 140^\circ = 90^\circ

The back bearing from

N

M

230^\circ

. Subtracting the forward bearing to

P

(

140^\circ

) gives the enclosed interior angle.

Apply the Pythagorean theorem to right-angled triangle

MNP

MP = \sqrt{MN^2 + NP^2} = \sqrt{24^2 + 10^2} = \sqrt{576 + 100} = \sqrt{676} = 26\text{ km}

Because

\angle MNP = 90^\circ

, triangle

MNP

is a right-angled triangle with hypotenuse

MP

Key Concept

Bearings and Right-Angled Triangles

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