A hiker starts at camp C and walks 5\text{ km} due East to checkpoint A. From checkpoint A… — JAMB UTME

A hiker starts at camp $C$ and walks $5\text{ km}$ due East to checkpoint $A$ . From checkpoint $A$ , the hiker then walks $3\text{ km}$ on a bearing of $150^\circ$ to reach checkpoint $B$ . What is the direct distance from camp $C$ to checkpoint $B$ ?

$7\text{ km}$ Answer
B
$\sqrt{34}\text{ km}$
C
$\sqrt{19}\text{ km}$
D
$8\text{ km}$

Answer

7\text{ km}

The interior angle at checkpoint

A

is calculated using three-figure bearings as

270^\circ - 150^\circ = 120^\circ

. Applying the Cosine Rule

CB^2 = 5^2 + 3^2 - 2(5)(3)\cos(120^\circ)

yields

25 + 9 + 15 = 49

. Taking the square root gives the direct distance of

7\text{ km}

Step-by-Step Solution

Determine the interior angle

\angle CAB

at checkpoint

A

\angle CAB = 120^\circ

Due East corresponds to a bearing of

090^\circ

. Coming into

A

from

C

means line

AC

points West (

270^\circ

). The bearing of

B

from

A

150^\circ

. The interior angle between vector

AC

pointing West (

270^\circ

) and vector

AB

on bearing

150^\circ

270^\circ - 150^\circ = 120^\circ

Apply the Cosine Rule to find length

CB

CB^2 = CA^2 + AB^2 - 2(CA)(AB)\cos(\angle CAB)

We have two sides (

CA = 5\text{ km}

AB = 3\text{ km}

) and the included angle (

\angle CAB = 120^\circ

Substitute the known values into the Cosine Rule formula

CB^2 = 5^2 + 3^2 - 2(5)(3)\cos(120^\circ) = 25 + 9 - 30(-0.5) = 34 + 15 = 49

Since

\cos(120^\circ) = -\cos(60^\circ) = -0.5

, the negative sign inside the cosine product cancels with the subtraction sign in the formula.

Take the positive square root to find the distance

CB = \sqrt{49} = 7\text{ km}

Distance must be positive.

Key Concept

Cosine Rule for non-right triangles in bearings problems

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