Lesson video

Hello, I'm Mrs. Lashley, and I'm looking forward to working with you as we go through this lesson.

I really hope you're looking forward to it and you're ready to try your best.

So our outcome today is to be able to use our knowledge of loci and constructions to solve problems. On the screen, there are some keywords that you've met before in your learning and I'm gonna be using during the lesson, so I suggest you pause the video, just reread them and make sure you feel familiar before we make a start.

So this lesson on advanced problem solving with loci and constructions is split into two learning cycles.

The first learning cycle is loci, constructions and bearings, and the second learning cycle is loci, constructions and areas.

So we're gonna make a start on the loci, constructions and bearings.

A boat is leaving a port on a bearing of 130 degrees.

Laura says, "Using a protractor and measuring from North, we can locate the boat." So she's got her protractor, and she's measuring clockwise 'cause bearings are measured clockwise.

She's made a mark at 130 degrees, and that's where she thinks that the boat is.

Lucas says, "That is only a possible location for the boat." So why is Lucas correct? Have a think about that.

So Lucas is correct because actually, the boat could be anywhere along the ray line.

We know it's on a bearing from the port, but we don't have any distances from the port to give it an exact location.

Laura recognises what her mistake was and says, "Oh, it's the locus of points but not the exact location." So more information is needed to find the exact location.

Can you think about what information we could be given so that we would know exactly where the boat is in comparison to the port? So here's a check for you.

So what bearing is the boat travelling on if this is the locus of points that a boat is on having left the port? Pause the video, and when you're ready to check, press play.

So the boat is travelling on a bearing of 110 degrees.

So if we continue thinking about what other information do we need in order to find the location of the boat, so here we've got some further information, which is the speed it is travelling.

So it's travelling at a speed of 40 kilometres per hour.

So where is the boat? Well, we know the boat is along the locus of points that is 130 degrees from the port, so along that ray line.

We have a scale, which is one square is 10 kilometres.

So where's the boat? Well, Lucas says, "Well, actually, the location will depend on how long it has travelled for." So just because we have the speed it is travelling still doesn't give us the boat's location.

So after one hour, the boat will have covered 40 kilometres.

That's what a speed of 40 kilometres per hour means.

And we can use our squares because that's our scale to know that it will be four squares away from the port, but not just four squares in any direction but along that ray line.

And we can use our pair of compasses to mark an arc that is four squares away from port B, that would be our radius would be four.

And where it intersects would be the location of the boat after one hour of travel.

If it was two hours of travel, then it would now be 80 kilometres from the port, and so we could draw an arc that has a radius of eight squares or 80 kilometres so that we know that moment in time, the boat would be located there.

So here's a check for you.

A boat leaves on a bearing of 55 degrees at a speed of 30 kilometres per hour.

How long has the boat been travelling for? Pause the video, and when you're ready to check, press play.

Well, it's travelled 60 kilometres, and we know it's travelled 60 kilometres 'cause we can see that the radius of that arc is six squares, which equates to 60 kilometres.

So 60 kilometres has travelled at a speed of 30 kilometres per hour.

So the distance has doubled, so therefore, the time it will have troubled too, so it's travelling for two hours.

The same boat that we've been trying to locate is on a bearing of 130 degrees from the port.

But another boat, which we're gonna call C, is due East of the port and picks up boat B on its radar system on a bearing of 235 degrees.

Where is boat B? So we can see the location of boat C.

And where is boat B? While we can draw out bearing.

The locus of points that are on a bearing of 130 degrees from point P, and the locus of points that are on a bearing of 235 degrees from boat C, where they intersect is where the boat would be.

So this further information allows us to find the exact location of boat B.

So here's a check.

Which of the following describes the location of the boat B? So you can see the accurate construction on the screen, and which of the descriptions matches it.

So pause the video, and then when you're ready to check, press play.

So description B.

And description A is very similar, but the two bearings are the wrong way round.

So it's important that we get those correct.

And part C, the bearing or the angle that's given is the angle between the two ray lines.

However, that is not a bearing because a bearing needs to be measured from North in a clockwise direction.

So we're gonna look at a different scenario for loci, constructions and bearings.

So this time, there are three pylons in a local area.

pylon B is at a bearing of 150 degrees from pylon A.

pylon B and C are both 500 metres from pylon A.

pylon C is due West of pylon B.

And a diagram may be helpful here.

You can see that there is a scale in the top right corner, so that distance, that length represents 100 metres.

So I'm gonna start with the first information point.

But you don't need to start with the first information point, but that's the one I'm gonna start with.

So I'm gonna place pylon A on my diagram, and then I'm going to find the location of pylon B and C relative to that point.

So pylon B is at a bearing of 150 degrees from pylon A.

So I can measure that angle of 150 degrees, and I know that that doesn't give me the exact location of pylon B but it tells me a locus of points that pylon B could be.

Then I can move to the second point, and that says that pylon B and C are 500 metres from pylon A.

Pylon A is a fixed point, so think about what the locus of points looks like when it's a fixed point.

And this one I could have done first before my bearing.

So I need to set up my pair of compasses to 500 metres, and using that scale, so it'll be five times that length, and I can draw my locus of points.

So all points along the circumference of that circle are 500 metres from pylon A.

Because I know that pylon B is along the ray line somewhere and I also know it needs to be 500 metres from pylon A, the point of intersection of those two locus is where pylon B is located.

Now I'm on to the final information point, which is pylon C is due West of pylon B.

And if we think about our compass directions, North and West are perpendicular directions.

So we now need to think about how we can construct a perpendicular line that passes through B that is perpendicular to the North.

So we're gonna do that by thinking about perpendicular bisectors, the skill, the knowledge we have.

So if I extend the North line, so that is still in the direction of North, I've just extended it, I now want a perpendicular line that passes through B.

So think about how we construct that.

So the steps are we're gonna create a line segment where B, our point off of the line, is equidistant to either end.

So we're gonna use pair of compasses to do that because that's a fixed distance.

That distance is your choice.

So you can see my arc here that crosses through the North line.

And now, I'm gonna do a perpendicular bisector on the line segment between those two intersection points.

So I would be putting my pair of compasses at one of those arcs which meets the North line and drawing two arcs.

And without changing the radius of my pair of compasses, do it from the other intersection point and draw my line.

So I've got a perpendicular bisector between that line segment that I created.

I created it such that the point B was equidistant from both ends of it.

So now I know where pylon C is.

It is where the perpendicular bisector or the perpendicular line that passes through B also intersects with the circle.

So a road has been planned that runs equidistant to pylon A and B.

Will pylon C need to be moved? So remember, these are electricity pylons in a town, and we now know that a road is gonna run equidistant from pylon A and pylon B.

So if we remove our construction lines just so we've got our locations only, we need to find the road, and we know it's equidistant to A and B.

So what does that mean? Well, it means that this needs a perpendicular bisector because a perpendicular bisector gives us the locus of points that is equidistant from both A and B.

So I'm gonna put a line segment, a faint line segment on and construct a perpendicular bisector.

And if you draw the perpendicular bisector, which indicates our road, the road would actually be passing through point C, which are pylon C.

So yes, the pylon will need to be moved, or they will need to reroute the road.

So here's a check for you.

So for this check, you need to label the pylons A, B, and C given the information on the screen.

So pause the video, read through it, and when you're ready to check, press play.

So C, then A, then B.

So going back to ships or boats, things in the water, a ship needs to pass through a point that is midway between two buoys.

So the buoys are B1 and B2, and the ship is the S.

Aisha and Jun both do some constructions to find the correct point to pass through.

So you can see their constructions.

And now, I want you to think about who has constructed the correct bisector.

So Aisha has done an angle bisector, and Jun has done a perpendicular bisector.

So who has chosen the correct bisector for this scenario? Jun has, and that's because it is two single points.

So the equidistant line of two points is a perpendicular bisector.

Aisha has constructed an angle bisector where she has joined the three points together with line segments, which isn't necessary in this scenario.

So continuing with this, using Jun's correct construction, what bearing should the ship travel? So you can see the ship, that's the one with the S, it needs to pass through the points that is midway between the two buoys.

So what's the bearing? So where is that point? That construction allows us to find that point.

Which point is it? Well, it's the point that's halfway, the bisector part of the two buoys, the midway.

So what's that bearing? Well, we're gonna measure using a protractor.

And in practical terms, you're probably gonna measure the obtuse angle and subtract from 360 degrees to get the reflex bearing.

But do remember that the bearing needs to be measured in the clockwise direction.

So 246 degrees is the bearing that the ship would need to travel from the location it's at in order to pass through the midway point of the two buoys.

So here's a check for you.

Which point represents the location of the point midway between the points A and B.

So there are four points that you could choose from, but which one is the location of the midway? Pause the video, and then when you're ready to check, press play.

So I'm hoping you went for D, and that's because of the perpendicular bisector allows us to bisect this line segment between A and B.

So that finds the midpoint of the line.

So we're now on to the first task of the lesson.

So question one, a yacht is sailing on a bearing of 80 degrees from a port.

Draw the locus of the yacht's path is part A.

And then you're told the speed, and it's travelled for 30 minutes, so locate the yacht.

And then for part C, after that 30 minutes, another yacht is 10 miles away on a bearing of 220 degrees from the first yacht, so mark its location.

So pause the video, and then when you press play, we'll move to question two.

Here's question two, and this is about three phone masts in a local area.

There are four information points.

You need to use those to find the location of mast A and mast C.

So pause the video, and when you're finished with that, press play and we'll move to question three.

Here's question three.

So the scenario here is a remote control boat race is taking place.

We can see the start and the finish.

The boat needs to pass through the checkpoints, which are pairs of flags, such that it's equidistant to both.

On the diagram, you can see C1 and C1.

So that's checkpoint one, and they're the pair of flags that go together.

And then you've got checkpoint two, and they're another pair of flags.

So for part A, you need to construct the appropriate bisectors to show the boat's path through the course checkpoints.

And then for part B, you need to write down the bearing of the finish point from the start point.

So pause the video.

When you finish, we're gonna go through the answers to task A.

So here's the answers to question one.

All the constructions are on the diagram, and hopefully, yours looks very similar to this.

So part A, you needed to draw the locus of the yacht's path.

So that was you measuring the bearing 80 degrees from the port, so using the North line to get your 80-degree angle, and drawing a ray line.

Then for part B, you were told the yacht was travelling at seven miles per hour, and you needed to show the location of the yacht after 30 minutes.

So you need to think about 30 minutes as half an hour, and therefore, the distance travelled would be 3.

5 miles.

Using the scale, that would be 3.

5 squares.

So you could draw a circle with a radius of 3.

5 squares and look at where it intersects your ray line from part A, and that would be the location of the yacht after 30 minutes.

For part C, it was after 30 minutes, so we're using the location of A.

But there is this second yacht that we're gonna call B.

And we know it's 10 miles away, and it's on a bearing of 220 degrees, so mark it's location.

So it doesn't matter which order you did these two constructions.

You needed to draw a circle.

We can only see an arc part of the circumference on the diagram, but that needed to be 10 squares of a radius because it was 10 miles.

And then you also needed to measure the bearing, 220 degrees.

And then when those two intersect would be where the location of the yacht B is.

Question two, you needed to read through the four information points, decide the appropriate order in order to locate A and C.

B was already given to you.

So using the first two information points, you can say that mast A is on a bearing of 140 degrees, so we can measure that angle.

And the ray line indicates where mast A will be located somewhere along it.

The same for the second point.

Mast C is on a bearing of 50 degrees.

So again, the 50-degree angle is overlapping with 140 degree.

The angle between the two ray lines is 90.

Then we've got mast A is 1,100 metres from mast C, and mast B is 900 metres from mast C.

So the third point, you can't make use of at this point because you don't know where A is and you don't know where C is.

But the fourth point includes mast B, and we do know where mast B is located.

So mast B is 900 metres from mast C.

It's the same as saying mast C is 900 metres from mast B.

So we can use the location of B, draw our locus of points that are 900 metres, which would be nine squares using the scale, and where that intersects with the bearing is where mast C is.

So that's how you could have found where mast C is located.

And now that you have mast C, you know that the distance between C and A is 1,100 metres, so you can draw your locus of points that is 1,100 metres from C.

And where that intersects with a bearing of 140 degrees from B will be the location of A.

On to question three.

You needed to construct the appropriate bisectors.

In both cases, it was perpendicular bisectors.

So you needed to use perpendicular bisectors for checkpoint 1 and a perpendicular bisector for checkpoint 2.

And then basically, you've shown the route that the boat needs to take.

It will start at the start.

It will travel equidistant through the checkpoint 1.

And at the point at which that perpendicular bisector meets the other perpendicular bisector is where the boat would need to spin round, change its direction of travel so that it now travels along the perpendicular bisector through checkpoint 2, keeping it equidistant from both flags, and then that takes it to the finish line.

For part B, you needed to write down the bearing of the finish from the start.

So if we draw a line segment between the start and finish, a North line in order to measure the bearing, use our protractor, that's 153 degrees.

So we're now up to the second learning cycle, which is loci, constructions and areas.

So three motion sensors are set up for security.

So the rectangle represents the area of land that is being protected, and we've got three sensors.

So sensor 1 has a range of 15 metres, sensor 2 has a range of 12 metres, and sensor 3 has a range of 20 metres.

And you can see where the sensors are located.

What area does sensor 1 cover? Well, sensor 1 is just a point, so we're gonna draw a circle for our locus, and so it's the locus of a single point.

Some of the circle is external to the area we're trying to protect, so we've only got part of a circle that we can actually see.

But we'd use our pair of compasses, and we'd fix the radius using the scale and make sure it's got a a radius of 15 metres.

What about the area for sensor 2? Well, that's gonna, again, this one's a segment because part of it's being cut by a cord.

But using the scale, we're gonna draw our circle because it's a locus of a single point.

And lastly, what about sensor 3? Well, similar idea, that sensor 3 is a single point, so we need to show the region of points that are up to and include in 20 metres from that single point, so we'd use our pair of compasses.

What areas are not covered by the sensors? Well, it's gonna be the regions that are outside of range for all of the sensors.

The circles show the boundary where the range, they go out of range, so outside of all three of them.

So here's a check for you.

Sensor 1 and sensor 2 are activated, so they have picked up motion.

Where might the intruder be? So pause the video.

When you're ready to check, press play.

So it's gonna be this region here, and that's the area that is covered by both sensor 1 and the sensor 2 only.

So moving on and continue with thinking about loci, constructions and areas, we've got a scenario here where a dog is tied to the wall of an outbuilding.

The outbuilding is five metres square, and the lead is eight metres long.

The region that the dog can go is shown here.

How far along the wall is the dog tethered? So that cross is where the dog is tied.

How far along that wall is it? I'm gonna use this part of the diagram to help me get to an answer.

And so I know that the outbuilding is five metres square, and the dog lead would be tethered at that cross.

When it got to that corner and turns around the corner, it gets five metres.

And the lead is eight metres in length, which means that this part here would be three metres in order to total eight.

So we could describe it as three metres from the left corner or two metres from the right corner depending on how you described it.

So how much space can the dog reach? So we can see that this is the region it can go, but how much space is that? So what's its actual area? Well, we need to total the semicircles, the quarter circles to find that area.

This is a compound shape made of different sectors.

So let's look at this a bit closer.

So if we start with the semicircle, this is when the lead is at maximum length of eight metres.

And it's not caught along or going around the corner, so it's on this one side of the building.

So it's a semicircle.

So it's 180 out of 360 degrees, times pi, times eight squared, which is a half times pie times eight squared.

So the half is, what fraction of the full circle is it using the angle between the two radii? And that simplifies to 32 pi.

Now, if we look at this quarter circle, it has a radius of five metres.

We know that because once the dog gets to that corner and turns around that side of the outbuilding, the lead would get trapped.

Three metres of it would be taut to the wall, and therefore, there's only five metres of free lead left.

So we need one quarter of a full circle with a radius of five.

Why one quarter? Well, because the angle between the two radii is 90 degrees, and that's out of 360 degrees for a full circle.

So one quarter of pi times five squared simplifies to 25 over four pi square metres.

We're leaving it in terms of pi so that we don't have any rounding errors as we work through the solution, especially because there's multiple calculations.

Now, if you look at this quarter circle, well, two metres of the lead would take us from the tether point to the corner, and that leaves six metres that can wrap around that side of the building.

So it's a quarter circle with a radius of six.

So one quarter of pie times six squared simplifies to nine pi.

Lastly, we've got this final quarter circle.

And so if two metres of the lead gets us to the corner, and then the outbuilding is five metres long, when it turns around that corner, there's only one metre of lead left.

A quarter circle of radius, one metre simplifies to a quarter pi.

So what space can the dog reach? Well, it's the total of all four parts, and so that comes to 95 over two pi.

As an exact answer, square metres, 149.

2 square metres to one decimal place, or you could say 149 square metres to three significant figures.

So here's a check.

A dog is tied to the wall of an outbuilding, so the same scenario.

The region that the dog can go is shown here.

What are the missing lengths? So pause the video, and then when you're ready to check, press play.

So A would be 12 foot, so that's the length of the lead or the rope that this dog is tied to.

And you can get that using the three foot and the nine foot.

B would also be nine foot.

So because it's a radius, it's a fixed length throughout the sector, so nine foot.

C is 7.

5 feet.

There's a couple of ways you could do this.

You could use the 4.

5 feet and think about how much more lead there would be to get to 12 feet.

Alternatively, it's the length of the outbuilding.

And you can see that the top edge is 4.

5 and 3 feet, which has a total of 7.

5 feet.

And lastly, there'd be 1.

5 feet of the lead that could wrap around to that third edge.

So if the lead has gone 3 feet to the corner along the 7.

5 feet of the building, that's 10.

5 feet of the lead used, which leaves you 1.

5 feet.

So here are the areas in square feet of the four parts to the locus.

What is the total area? So pause the video, and then when you're ready to check, press play.

So 855 over eight pi square feet.

Here, we're looking at a rear windscreen wiper that cleans only part of the glass.

So maybe you've recognised that when you've been in a car or you've seen a car.

The wiper blade is 24 centimetres long and pivots 165 degrees from the horizontal.

What region of the screen gets wiped? So you can see that it creates a sector.

If the rear screen has an area of 2,160 square centimetres, what area does not get cleaned by the wiper blade? So the wiper blade only cleans a part of the glass and it cleans a sector.

If we have the full area, how do we get the area that's not wiped? Well, the region is a sector, as I say, so we can work out that area by doing 165 out of 360 'cause that's what fraction of a full circle this would be.

The radius is 24, so 264 pi square centimetres.

So the area that is not cleaned is the bit that's not wiped in this diagram, and we can do that via subtraction.

So here's a check for you.

The windscreen has an area of 2,500 square centimetres.

The wiper blade is 30 centimetres long and pivots 170 degrees from the horizontal.

Work out the area of the glass that is wiped to one decimal place.

Pause the video, and when you're ready to check, press play.

So it'll be 1,335.

2 square centimetres.

Is less than or more than half of the windscreen wiped by the wiper blade? Pause the video, and when you're ready to check, press play.

So more than half is wiped because if the area is 2,500 square centimetres, half of that is 1,250, and 1335.

2 is more than 1,250.

So Alex wants to work out the area of the triangle ABC.

He says there are two ways to calculate the area of a triangle, using half ab sin C or half base times perpendicular height.

Without knowing an angle or having a protractor, I'll need to calculate it.

So this triangle here, ABC, at the moment, we haven't got any values, any dimensions to it, and Alex has said he doesn't know an angle, and he hasn't got a protractor.

So how can Alex calculate the angle using the cosine rule? So he measures the three edges.

So there, we can see their measurements.

And Alex says, "I need to use a rearrangement of the cosine rule to calculate the angle A." So here is the rearrangement of the cosine rule.

So he substituted in his b, his c, and his a.

He substituted that in, evaluating that cos A is equal to minus 901 over 2,405, or as a decimal form, it's a recurring decimal.

How do we get from cosine of A while we need to do arc cos? And that comes out as 112 degrees.

So we now know that the angle A that's opposite the length A is 112 degrees.

So now we can calculate the area of the triangle.

So area of a triangle is half ab sin C, or half bc sin A, or half ac sin B.

Which of the equivalent forms will Alex use? So pause the video and think about that.

So he's gonna need to use the middle one, which involves the angle A because that's the angle he has calculated.

So the area is half times 6.

5, times 3.

7, times sin of 112.

Using the calculator, that's 11.

1 square centimetres to three significant figures.

So Alex says, "I think I could have also used constructions to find the area using half times base times perpendicular height." So what constructions could Alex be talking about? So the altitude of a triangle is equal to the perpendicular height, so I can construct and measure an altitude.

I will construct the altitude that passes through vertex A by considering it as a perpendicular bisector to a point off of the line segment BC.

So he's gonna go through his construction if we just focus on the line BC and that point off of it.

So the construction steps is to make a line segment such that A is equidistant from both ends of that line segment.

And the easiest way to do that is to use an arc because then you know that that distance from A to the intersection and A to the intersection is equal.

Then we're gonna do a perpendicular bisector using those two points of intersection as the ends of the line segment, so making sure that our pair of compasses is over half the distance so that they do intersect.

And we get this shape that we're quite familiar with, I'm sure, and we only need the sort of arc that intersect, and we can now draw on our perpendicular bisector.

So this is now a perpendicular line that passes through the point A.

We only really care because of the context for the part that goes from A to the line and not past it.

Alex continues to say, "Now that I've constructed and measured this altitude," so he's measured it to 2.

6 centimetres, "I can calculate the area using half times base times perpendicular height." So the altitude he constructed was perpendicular to 8.

6, so now he's gonna do base and perpendicular height, and it comes out as 11.

18 square centimetres.

So why are the two methods returning different answers? Have a think about that.

So the first one was 11.

149, rounded to 11.

1 to three significant figures, whereas this one's 11.

18, which would round to 11.

2 to three significant figures.

So why are the two methods returning different answers? So there's inaccuracy caused by the measuring.

A longer altitude would have less percentage error and might give a closer answer to the other method.

So here is a check for you.

Constructing the perpendicular bisector of an edge will give you the perpendicular height of the triangle.

True or false? And justify your answer.

The altitude of an isosceles triangle does bisect the edge it meets.

The altitude is always perpendicular to the edge it meets but does not necessarily bisect it.

Pause the video, and when you're ready to check, press play.

So it's false, and that's because of B.

So for some triangles, the altitude is going to bisect the edge that it meets, but not for all.

So we're up to the last task of the lesson.

So here is question one.

Press pause, read through it.

And when you're ready for the next question, press play.

So here's question two.

Once again, pause the video, read through it, have a try, and when you're ready for question three, press play.

Question three is on the screen now.

Make sure you read through it.

You might want to sketch yourself a diagram before you start.

So press pause, have a try.

And then when you press play, you're gonna have the last question of this task.

And here is question four where you need to work out the area of triangle ABC that you can see on the screen three different ways.

So pause the video.

And when you finish with question four and press play, we're gonna go through the answers to Task B.

So question one, you needed to firstly draw the locus of each sensor using the scale, then you needed to shade specific regions.

So the sensors' loci are shown here.

Shade the area that's covered by all three sensors, so that's the part that is within each circle.

That would be this part here.

Different colour, shade the area that is not covered by any of the sensors.

So there's four regions that you should have shaded.

And for part C, a different colour, shade the area that is only covered by sensor 2.

So it's within the locus of sensor 2 but not within the other two sensors, and that would be this region.

Question 2, the shape of the lawn is on the screen there.

So it was a garden pergola that had a footprint of a regular hexagon where each edge was 1.

8 metres.

The lawn is designed to be the shape of a locus made by a rope of 4.

8 metres, fixed a third of the way on one side of the pergola.

So it was lots of sectors.

There was a semicircle, and then there were five further sectors.

Part B was work out the area of the lawn.

So the semicircle is half of a full circle.

And then the other sectors were 1/6, and that was used in the exterior angle of a regular hexagon to know that that would be 60 degrees, that internal angle.

The radii of each of those sectors was dependent on how much rope was gonna wrap further round.

The radii were 4.

8, 3.

6, 1.

8, 4.

2, 2.

4, and 0.

6, and that comes out as 57.

1 square metres to three significant figures.

Question three, part A, what area of the rear windscreen is wiped by the wiper blade? So the locus is a sector.

It's got a radius of 28 centimetres and an angle of 150 degrees.

Because the windscreen was given in metres, it made sense to change the wiper blade length to metres.

Otherwise, we'd need to change the units at a later point.

So 150 out of 360, times by pi, times by 0.

28 squared gives us 49 over 1,500 pi.

For part B, what area of the rear windscreen is not wiped by the wiper blade? So you needed to work out the air of the windscreen.

And because you were told it was a trapezium in shape, we could use the formula for the area of a trapezium.

And then subtract the area of the sector that you worked out in part A, which leaves 0.

72 square metres to two decimal places.

Question four, you needed to work out the area of the triangle in three different methods.

So for part A, it was measure angle C and then use half ab sin c.

So using your protractor, so 19.

1 square centimetres was the area to three significant figures.

For part B, you needed to calculate the angle A and then use an appropriate formula.

So this would be a bit like when Alex did it, so using the cosine rule to calculate the angle.

Try and use the most exact angle, so leaving it on your calculator, using the answer key.

So the area was 19.

1 square centimetres to three significant figures.

And then lastly, for part C, constructing and measuring the altitude that passes through B.

So you can see the construction there, and the green part is the altitude that we actually needed to measure.

So that measures to 7.

6.

When we use half times base times perpendicular height, then we get 19.

2 square centimetres.

So to summarise today's lesson on advanced problem solving with loci and constructions, constructions such as perpendicular bisectors and angle bisectors can sometimes be used in problems that also involve bearings.

And problems that involve area can sometimes require loci to identify the correct region in question.

So well done today, and I look forward to working with you again in the future.