Lesson video

Hi, everyone, my name is Ms Ku, and I'm really happy to be joining you today, because today we're looking at bearings, a great topic as there are so many real-life applications.

Let's make a start.

I hope you enjoy the lesson.

Hi everyone, and welcome to this lesson on problem solving with bearings under the unit "Bearings." By the end of the lesson, you'll be able to use your knowledge of bearings to solve problems. Today's lesson will consist of a lot of keywords.

Clearly, we'll also be looking at the keyword "bearing," and a bearing is an angle measured in degrees from north in the clockwise direction and written with three figures.

For example, the angle of 82 degrees is written and said as 082 degrees.

Today's lesson will be broken into three parts.

We'll look at bearings with three or more positions first, then bearings and right-angled triangles, and then further problems involving bearings.

So let's make a start looking at bearings with three or more positions.

Now, looking at any compass, you'll see east is always 90 degrees clockwise from the north.

Therefore, whenever due east is referenced, it means 90 degrees clockwise from the north.

But what bearing is due west? Due west is a bearing of 270 degrees.

This is because it's always 270 degrees clockwise from our north.

So let's have a look at a quick check.

What is the bearing when a position is due south? What is the bearing when a position is northeast, and what is the bearing when a position is southwest? See if you can give it a go.

Well done.

Well, hopefully you spotted that the bearing when a position is due south is 180 degrees.

Next, what is the bearing when a position is northeast? Well, it's 045 degrees.

Next, what is the bearing when a position is southwest? Well, it's 225 degrees.

Really well done if you got this.

Now, three towns form an equilateral triangle.

C is due east of A, and we're asked to work out the bearing from A to B.

Laura says, no angles are given, so I'm going to use my protractor.

Explain why Laura is incorrect when she said the bearing from A to B is 033 degrees.

Have a little think.

Well, it's because the diagram's not drawn accurately, and this means we must calculate and not measure.

So, given the fact that we know no angles are given, we are given enough information to work out the answer.

This is because we know the triangle is equilateral.

Therefore we know each interior angle.

Each interior angle is 60 degrees.

Now, it states that C is due east of A.

This means we know C is 90 degrees from the north at A.

So I'm gonna indicate it here.

Now we can work out the bearing from A to B.

I've indicated it in purple here.

What do you think that bearing is? Well, the bearing from A to B is 90 degrees.

Subtract 60 degrees, which is 030 degrees.

Now let's have a look at a check.

Three towns form an equilateral triangle.

C is due east of A.

I want you to work on the bearing from B to A, and I want you to work out the bearing from C to A.

Well done.

Let's see how you got on.

Well, we know the interior angle of an equate triangle is 60 degrees, so therefore I've indicated it on our diagram.

We also know that C is due east of A.

So I've indicated my 90-degree angle here.

Now I know this angle is 30 degrees, and I can use my knowledge of co-interior angles to work out this angle to be 150 degrees.

The question wants me to work out the bearing from B to A, so it's simply 360 degrees, subtract 150 degrees, giving me a bearing of 210 degrees.

Next, let's work out the bearing from C to A, which is this one.

Well, I know the anticlockwise angle is simply 90 degrees, so that means 360 subtract 90 gives me a bearing of 270 degrees.

For more complex questions, no diagrams are given.

That means we need to sketch one.

So here's a question.

It says, three cities, A, B, and C, are around the UK, such that C is due east of city B.

City B is on a bearing of 074 degrees from city A.

The distance between cities A and B is the same as the distance between cities B and C.

We're asked to draw a sketch of these three cities and label all the relevant information.

Part B wants us to work out the bearing from city C to city A.

So, let's make a start by doing our sketch.

Reading one sentence at a time, draw what is given.

We know C is due east of city B, so I've indicated it here.

We also know city B is on a bearing of 074 degrees from city A.

So I've indicated it here, as well as my north.

Next we know the distance between cities A and B is the same between cities B and C.

So, using hash marks we can indicate equal lengths.

Now let's indicate the north for each city, and let's see if we can work out the bearing from city C to city A.

Now our sketch is complete.

Well, drawing this line segment shows the bearing we need to calculate.

So now we know what we need to calculate, let's use our knowledge of co-interior angles.

We know this angle must be 106 degrees.

We know sum of angles around a point is 360 degrees, so I can work this out to be 164 degrees.

Now, you might spot we have an isosceles triangle, so we can work out the equal unknown angles.

They will be eight degrees.

Knowing that the sum of angles around a point is 360 degrees, I have now worked out my bearing from city C to city A.

It's simply 360 degrees.

Subtract my 90, subtract my eight, giving me a bearing of 262 degrees.

Now it's time for your check.

Using squared paper where the length of one square is 10 metres, I want you to draw the following.

Point A and B are 60 metres apart.

Point B is on a bearing of 090 degrees from point A.

Point C is 60 metres north of point B and on a bearing of 045 degrees from point A.

See if you can draw this on some squared paper.

Well done.

Let's see how you got on.

Well, you can really position A anywhere.

I've placed it here and I've identified my north.

We know B is 60 metres from A, and we know B is at a bearing of 090 from point A.

So I've positioned it here.

Now we know C is 60 metres north from B and a bearing of 045 degrees from A, so that means C must be here.

Now, using our diagram, I want you to work out the following bearings, the bearing from C to A and the bearing from B to A.

Then D is the midpoint of AC, and I want you to work out the bearing from A to D, the bearing from C to D and the bearing from D to B.

Great question.

Take your time.

Well done! Let's see how you got on.

Well, let's work out the bearing from C to A, which is given here.

We know we need to work this out, but to do this, let's use our knowledge of co-interior angles.

So this angle must be 135, giving me the bearing from C to A to be 225 degrees.

The bearing from B to A must be found by using this line segment, and I've indicated the bearing here, so it must be 270 degrees.

Now we know D is the midpoint of AC.

So what's the bearing from A to D? Well, it's 045 degrees.

Position D and C are on that same line segment.

The bearing from C to D, well, it's 225 degrees.

Once again the positions A, D, and C are all on the same line segment.

Next, the bearing from D to B can be found by identifying this line segment, this angle, and then using our knowledge of co-interior angles, we know the bearing would be 135 degrees.

Great question.

Fantastic work, everybody.

Now it's time for your task.

Read the questions carefully, annotate anything you need.

Press Pause as you'll need more time.

Well done.

Let's move on to question two.

Fantastic question, use the grid carefully and pay attention to that scale.

Press Pause as you'll need more time.

Well done.

Let's move on to question three.

Another wonderful question.

Read the question carefully, and pay attention to that scale.

Well done.

Let's move on to these answers.

Well, for question one, drawing my diagram, I've indicated all the angles I need to help me work out these bearings.

So that means the bearing from X to Y is 030 degrees.

The bearing from Y to Z is 150 degrees, and the bearing from Z to X is 270 degrees.

Well done if you got this.

For question two, I've drawn my diagram, indicated my positions, and here are my bearings.

Very well done if you got this.

Next for question three.

Great question.

Here are my positions with my naught.

So that means the bearing of H from K is 045 degrees.

Well done if you got this.

Great work, everybody.

So now it's time for the second part of our lesson, bearings and right-angled triangles.

A yacht starts at a port and it's given the instruction to sail 60 kilometres due east and then a hundred kilometres due south.

What is the bearing from the port to the final position of the yacht? Well, let's start by sketching.

Here we have a yacht and I've indicated the north.

We know the yacht travels 60 kilometres due east.

So I've indicated it here, and then the yacht travels a hundred kilometres due south.

So I've indicated it here.

This is the bearing we need to work out.

So we know this is 90 degrees as the yacht travelled due east, but how do we find this angle? Have another think.

Well, we use trigonometry, but what trigonometric ratio do we use to work out this angle? Its tangent.

So let's substitute what we can to work out that angle theta.

We know tan theta is equal to a hundred over 60, so I can work out my angle to be 59.

0 degrees to one decimal place.

So now I know this angle is 59.

0 degrees to one decimal place.

We know the bearing from the port to the yacht is 90 plus 59.

0, giving me a bearing with three digits to be 149 degrees.

Well done.

So let's have a look at a check.

A yacht leaves a port and travels 30 kilometres due east and then 20 kilometres due south.

I want you to work out the bearing from the port to the yacht and then work out the bearing from the yacht to the port.

See if you can give it a go.

Press Pause if you need more time.

Well done.

Let's see how you got on.

Well, here is my diagram, and from my diagram you might notice I need to work out that angle theta.

To do this I'm going to use tan again.

So tan theta is equal to 20 over 30, working out my angle of theta to be 33.

7 to one decimal place.

So that means the bearing from the port to the yacht is 90 add our 33.

7, which gives us a bearing of 124 degrees.

Remember bearings have to have three figures.

Well done if you got this.

Moving on to part B, we're going to use our knowledge of co-interior angles, and the bearing from the yacht to the port is 360 degrees.

Subtract 056 degrees giving me a bearing of 304 degrees.

Well done if you got this.

Let's have a look at a check.

Sophia was given this question and she's drawn this diagram and given her answer.

Unfortunately she's made a mistake.

Can you spot it? The question says, a yacht starts at a port.

It's given the instruction to sail 10 kilometres due east and then 30 kilometres due north.

And the question wants us to work out the bearing from the port to the yacht.

Can you find out where Sophia's mistake is? Hopefully you spotted she's drawn her diagram incorrect.

So let's help Sophia and draw the correct diagram and work out the answer.

Press Pause as you'll need more time.

Well done.

Let's see how you got on.

Here's my diagram and my indicated lengths.

So I'm using tan again.

So tan theta is equal to 30 over 10.

Working this out gives angle of theta to be 71.

6 to one decimal place.

So I know this angle is 71.

6 degrees, but notice the question wants you to work out the bearing from the port to the yacht.

So, given the fact that we know the yacht sailed due east, we have to do 90 degrees, subtract 71.

6 degrees to give me a bearing of 018 degrees.

The diagram is so important as it helps you visualise what bearing you need to find.

Great work, everybody.

So now it's time for your task.

Question one states, a yacht starts at a port.

It is given the instruction to sail 40 kilometres due east and then 120 kilometres due north.

Part A says, what is the bearing from the port to the yacht? And part B says, what is the bearing from the yacht to the port? See if you can give it a go.

Press Pause if you need more time.

Well done.

Let's move on to question two.

A helicopter travels due south from a heliport for 30 kilometres.

It then travels due west and arrives at the heli station.

The straight line distance from the heliport to the heli station is 50 kilometres, but what is the bearing from the heliport to the heli station? See if you can give it a go.

Press Pause if you need more time.

Well done.

Let's move on to these answers.

Well, here's my diagram.

So that enables me to find the bearing to be 018 degrees.

Using my diagram, I can work up the bearing from the yacht to the port to be 198 degrees.

Fantastic if you got this.

Press Pause if you need more time to copy this down.

Well done.

Let's have a look at question two.

Here's my diagram and I'm using cos.

So working out that angle, I know the bearing is 233 degrees.

Press Pause if you need more time to copy this down.

Well done, everybody.

So now it's time for the third part of our lesson, further problem solving with bearings.

For the unscaled diagrams, I want you to work out the bearing from B to A, and I want you to show all your working out.

See if you can give it a go.

Press Pause for more time.

Well done.

Well, here's my working out, giving me a bearing of 214.

For the next one, here's my working out, giving me a bearing of 250.

Next one, here's my working out, giving me a bearing of 202.

What procedure is always the same when finding the bearing of A from B given the bearing of B from A? Have a little think.

Well, first of all, we subtract the angle from 180 degrees, and then we subtract this answer from our 360 degrees.

So let's rename the bearing X and see what the bearing of A from B is in terms of X.

So you can see it in my diagram.

Well, we know the co-interior angle is found by 180 degrees, subtract our X.

Then the bearing of A from B is 360 degrees.

Subtract that 180 degrees, subtract the X.

So, simplifying gives me 180 degrees, add X.

So, if we know the bearing of B from A, we can work out the bearing of A from B by simply adding 180 degrees to that known bearing.

Now it's time for a check.

The bearing of B from A is reflex, and it's labelled as X degrees.

We're asked to write the bearing of A from B in terms of X, and I want you to show all your working out.

See if you can give it a go.

Press Pause if you need more time.

Well done.

Let's see how you got on.

Well, we know the anti-clockwise angle will be found by 360 degrees, subtract our X.

Now, the bearing of A from B using our knowledge of co-interior angles is 180 degrees.

Subtract about 360 degrees, subtract our X.

Simplifying means we can work out the bearing of A from B to be X, subtract 180 degrees.

Great work if you got this.

Well done, everybody.

So let's move on to your task.

Given the statements, put a tick in the correct column, and if the statement is sometimes true, explain the conditions when it is sometimes true.

See if you can give it a go.

Press Pause if you need more time.

Well done.

Let's move on to question two.

If the bearing from A to B is X, where X is less than 180 degrees and the angle A, B, C is Y degrees where Y is less than 180 degrees, I want you to write the bearing from B to C in terms of X and Y in its simplest form.

And then I want you to write the bearing from C to B in terms of X and Y in its simplest form.

See if you can give it a go.

Press Pause if you need more time.

Well done.

Let's move on to question three.

If the bearing from B to C is X degrees where X is less than 180 degrees, and the angle A, B, C is Y degrees where Y is less than 180 degrees, I want you to write the bearing from C to B in terms of X in its simplest form.

And I want you to write the bearing from A to B in terms of X and Y in its simplest form.

See if you can give it a go.

Press Pause for more time.

Well done.

Let's move on to these answers.

Well, hopefully, you've got these answers here.

The only instance when it's sometimes true is when the bearing from A to B is X degrees, the bearing from B to A is 180 degrees, add X degrees.

This is only sometimes true when the bearing from A to B is acute or obtuse.

Next, the other time when it's sometimes true is when the bearing from A to B is X degrees, the bearing from B to A is X, subtract 180 degrees.

This is sometimes true only when the bearing from A to B is reflex.

For C, great question.

Notice how I've annotated on my diagram to work out these following answers.

Press Pause if you need more time to copy them down.

And for question three, same again.

Notice how I've annotated as much as I possibly can on my diagram to work out these answers.

Press Pause if you need more time to copy it down.

Great work, everybody.

So, in summary, knowing compass directions can make bearings calculations much easier.

And we can use trigonometry to work out bearings when right-angle triangles are formed.

And when diagrams are unscaled, we can calculate answers using angle facts.

Using these angle facts we can form algebraic expressions for bearings.

Really well done, everybody.

It was wonderful learning with you.