A vessel departs from a harbor H and sails 12\text{ km} on a bearing of 040^\circ to reach… — JAMB UTME

A vessel departs from a harbor $H$ and sails $12\text{ km}$ on a bearing of $040^\circ$ to reach point $A$ . From point $A$ , it changes course and sails $5\text{ km}$ on a bearing of $130^\circ$ to reach point $B$ . What is the bearing of point $B$ from harbor $H$ , to the nearest degree?

$063^\circ$ Answer
B
$073^\circ$
C
$017^\circ$
D
$107^\circ$

Answer

063^\circ

(or

063^\circ

to the nearest degree)

The navigation path forms a right-angled triangle at point

A

because the reverse bearing of

H

from

A

(

220^\circ

) differs from the bearing of

B

from

A

(

130^\circ

) by exactly

90^\circ

. Using the right triangle trigonometric ratio

\tan(\angle AHB) = \frac{5}{12}

, the angle

\angle AHB

is found to be approximately

22.62^\circ

. Adding this angle to the initial bearing of

040^\circ

yields

062.62^\circ

, which rounds to

063^\circ

Step-by-Step Solution

Determine the back bearing of

H

from

A

and calculate the interior angle at

A

Back bearing of

H

from

A = 040^\circ + 180^\circ = 220^\circ

. Interior angle

\angle HAB = 220^\circ - 130^\circ = 90^\circ

Knowing the directions of

AH

and

AB

allows us to find the interior angle of

\triangle HAB

at point

A

Calculate the interior angle

\angle AHB

using right-triangle trigonometry.

\tan(\angle AHB) = \frac{AB}{HA} = \frac{5}{12} \implies \angle AHB = \arctan\left(\frac{5}{12}\right) \approx 22.62^\circ.

Since

\triangle HAB

has a right angle at

A

, tangent relates the opposite side

AB = 5\text{ km}

to the adjacent side

HA = 12\text{ km}

Calculate the total bearing of

B

from

H

\text{Bearing} = 040^\circ + 22.62^\circ = 062.62^\circ \approx 063^\circ.

The line

HB

lies to the right of line

HA

, so the interior angle

\angle AHB

is added to the initial bearing of line

HA

(

040^\circ

Key Concept

Bearings and Right-Angled Triangles

Estimated Time:1m 30s

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