A ship departs from a port P and sails 10\text{ km} on a bearing of 060^\circ to reach a p… — JAMB UTME

A ship departs from a port $P$ and sails $10\text{ km}$ on a bearing of $060^\circ$ to reach a position $Q$ . From $Q$ , the ship changes course and sails $10\text{ km}$ on a bearing of $150^\circ$ to arrive at point $R$ . What is the bearing of point $R$ from point $P$ ?

A
$015^\circ$
$105^\circ$ Answer
C
$135^\circ$
D
$150^\circ$

Answer

The bearing of point

R

from point

P

105^\circ

The back bearing from position

Q

to port

P

240^\circ

. The difference between

240^\circ

and the new bearing of

150^\circ

yields an interior angle of

90^\circ

at vertex

Q

. Since both distances

PQ

and

QR

are equal to

10\text{ km}

, triangle

PQR

is a

45^\circ-45^\circ-90^\circ

isosceles right triangle. Adding

\angle QPR = 45^\circ

to the initial bearing of

060^\circ

yields a bearing of

105^\circ

for point

R

from point

P

Step-by-Step Solution

Determine the back bearing of

P

from

Q

The back bearing of

P

from

Q

060^\circ + 180^\circ = 240^\circ

To find the internal angle at

Q

, we need the direction of

P

relative to

Q

Calculate the interior angle

\angle PQR

\angle PQR = 240^\circ - 150^\circ = 90^\circ

The difference between the line back to

P

(

240^\circ

) and the line to

R

(

150^\circ

) forms the interior angle at

Q

Determine the properties of triangle

PQR

and angle

\angle QPR

Since

PQ = QR = 10\text{ km}

and

\angle PQR = 90^\circ

, triangle

PQR

is an isosceles right triangle, so

\angle QPR = 45^\circ

The two equal sides subtend equal acute angles in a right-angled triangle:

(180^\circ - 90^\circ) / 2 = 45^\circ

Calculate the total bearing of

R

from

P

Bearing of

R

from

P = 060^\circ + 45^\circ = 105^\circ

Point

R

lies clockwise relative to the segment

PQ

, so we add

\angle QPR

to the initial bearing of

PQ

Key Concept

Three-point bearing calculations using geometry of parallel lines and right-angled triangles.

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