18. Bearings - Cardiff Tutor Company

Question 1

The diagram below shows the bearing of $A$ from $B$ .

Find the bearing of $B$ from $A$ .

Select the correct answer from the list below:

A: $107^\circ$

B: $287^\circ$

C: $73^\circ$

D: $253^\circ$

CORRECT ANSWER: A: $107^\circ$

WORKED SOLUTION:

The bearing of $B$ from $A$ is the missing angle $x$ in the following diagram.

It is helpful to remember that the interior angles between two parallel lines add up to $180^\circ$ .

x+y=180^\circ

So, we need to find missing angle $y$ first, which we can do by remembering that angles around a point add up to $360^\circ$ .

$287^\circ+y=360^\circ$
$y=360^\circ-287^\circ$

y=73^\circ

We can now use our knowledge of interior angles to find $x$ .

$x+y=180^\circ$ $x+73^\circ =180^\circ$ $x =180^\circ -73^\circ$ $x=107^\circ$

Question 2

A plane flies from airport $X$ to airport $Y$ on a bearing of $210^\circ$ .

What would be the bearing for the plane’s return journey to $X$ from $Y$ ?

Select the correct answer from the list below:

A: $210^\circ$

B: $070^\circ$

C: $150^\circ$

D: $030^\circ$

CORRECT ANSWER: D: $030^\circ$

WORKED SOLUTION:

We know that angles around a point add up to $360^\circ$ .

$a+210^\circ =360^\circ$
$a =360^\circ-210^\circ$
$a =150^\circ$

And we know that interior angles must add up to $180^\circ$

$b+150^\circ =180^\circ$
$b=180^\circ-150^\circ$
$b=30^\circ$

So, the bearing to X from Y must be $030^\circ$

Question 3

The diagram below shows three points; A, B and C.

The bearing of B from A is $129^\circ$ , the bearing of C from B is $258^\circ$ , and the bearing of A from C is $059^\circ$ .

Find the missing angles inside the triangle ABC.

Select the correct answer from the list below:

A: $31^\circ, \hspace{1em}58^\circ, \hspace{1em} 91^\circ$

B: $21^\circ, \hspace{1em}76^\circ, \hspace{1em} 83^\circ$

C: $19^\circ, \hspace{1em}51^\circ, \hspace{1em} 110^\circ$

D: $53^\circ, \hspace{1em}62^\circ, \hspace{1em} 65^\circ$

CORRECT ANSWER: C: $19^\circ, \hspace{1em}51^\circ, \hspace{1em} 110^\circ$

WORKED SOLUTION:

We know that interior angles between parallel lines add up to to make $180^\circ$ , meaning we can fill in the angle between A and B.

$129^\circ+w=180^\circ$
$w=180^\circ-129^\circ$
$w=51^\circ$

We know that angles around a point add up to $360^\circ$ , so we can use that to find our first missing angle.

$258^\circ+51^\circ+x=360^\circ$
$309^\circ+x=360^\circ$
$x=360^\circ -309^\circ$
$x=51^\circ$

To find our missing angle $y$ , we can use our knowledge of interior angles again.

$51^\circ+51^\circ+59^\circ+y=180^\circ$
$161^\circ+y=180^\circ$
$y=180^\circ -161^\circ$
$y=19^\circ$

Finally, we can use our knowledge that angles in a triangle add up to $180^\circ$ to find missing angle $z$ .

$51^\circ+19^\circ+z=180^\circ$
$70^\circ+z=180^\circ$
$z=180^\circ -70^\circ+$
$z=110^\circ+$

Giving our us a triangle with angles:

19^\circ, \hspace{1em}51^\circ, \hspace{1em} 110^\circ

Question 4

For part of their Duke of Edinburgh award a group of students are taking part in an expedition.

For part of their expedition they walk on a bearing of $060^\circ$ for 3.5km. Using the scale 1cm : 1km, construct this part of their journey.

Select the correct answer from the list below:

CORRECT ANSWER: A

WORKED SOLUTION:

When drawing a diagram involving bearings, we always start by drawing our start point and the north line.

We then measure our angle clockwise from the north line, remembering that bearings are always three digits.

This is the students’ journey.

Question 5

A pilot is travelling between three airports, $A$ , $B$ and $C$ .

They fly from $A$ to $B$ to $C$ . The bearing of B from A is $030^\circ$ .

The bearing of $C$ from $B$ is $165^\circ$ . Airport B is 200km from Airport A, and Airport C 180km from Airport $B$ .

Draw the journey taken by this pilot using a scale of 40km:1cm.

Select the correct answer from the list below:

CORRECT ANSWER: C

WORKED SOLUTION:

When drawing a diagram involving bearings, we always start by drawing our start point and the north line.

We then measure our angle clockwise from the north line, remembering that bearings are always three digits.

After measuring the angle, the distance needs to be drawn correctly using the scale. Because the scale is 40:1, divide the km by 40 to find the length in cm.

200\div40=5

So, the line needs to be $5$ cm long.

The journey now goes from $B$ to $C$ , so a new north line needs to be drawn from $B$ .

Now that the north line has been drawn in the bearing of $165^\circ$ can be drawn in.

After measuring the angle, the distance needs to be drawn correctly using the scale. Because the scale is 40:1, divide the km by 40 to find the length in cm.

180\div40=4.5

So, the line needs to be $4.5$ cm long.