Lesson video

Hello everyone and welcome to this exciting lesson on circle theorems. I'm Mr. Gratton and get ready for this lesson where we will use our knowledge of a range of different circle theorems alongside trigonometry and Pythagoras' theorem or where ratios and algebraic expressions are involved.

Right, let's get started by having a quick look at some important keywords that might come in handy.

Pause now to check them out.

First up, let's have a look at how we can use Pythagoras' theorem and trigonometry alongside some circle theorems. Using Pythagoras' theorem or trigonometry alongside circle theorems can help us find angles, lengths, and areas of shapes on diagrams involving circles.

So for example, let's try and find the area of the triangle, A, B, C within that circle.

Pause here to think about or discuss which circle theorems can we use with this diagram and how will this help us to either use Pythagoras' theorem or trigonometry to find the area of that triangle, ABC.

We can use the alternate segment circle theorem to identify the angles BCD and CAB are both equal, both at 71 degrees.

Furthermore, AB is a diameter and therefore angle ACB is 90 degrees.

This is because of the circle theorem.

The angles at the circumference of a semicircle are always right angles or 90 degrees.

If we are looking to use either Pythagoras' theorem or trigonometry, keeping an eye on where the right angled triangles are located might come in handy.

Furthermore, labelling as much as possible can also sometimes come in handy.

So let's label angle ABC at 19 degrees because the interior angles of a triangle all sum to 180 degrees.

We can now focus on identifying lengths on triangle ABC.

Sometimes it helps to sketch a triangle on its own separate from the rest of the diagram in order to not get distracted by unnecessary measurement on a more complex diagram like this one.

Because the radius of the circle is five centimetres, the diameter AB is double that at 10 centimetres long.

This means that the hypotenuse of the triangle is also 10 centimetres.

Brilliant, we have the hypotenuse and a few angles so we can use trigonometry to find the length of one of the shorter sides of that triangle.

To find length AC, we can use the cosine ratio giving us a length of 3.

255 centimetres.

Length BC can be found using the sine ratio, giving us 9.

455 centimetres.

We can then use the exact values of these two perpendicular lengths on the triangle to find the area of this triangle at 15.

4 centimetres squared.

Great stuff.

Let's check your understanding.

We want to find the area of triangle AEF in multiple steps.

For step one, pause here to find the angle BCA and justify your answer with a circle theorem.

We have 77 degrees due to the alternate segment theorem.

Both angles are equal.

For step two pause here to find the angle ACF and justify your answer.

We have 35 degrees because angles about point C on a straight line, sum to 180 degrees.

Onto step three.

We can use the circle theorem, angles in the same segment are equal.

Pause here to find two angles that are equal to each other and then find the size of these two angles.

Both ACF and AEF are 35 degrees.

Onto step four.

Pause here to locate the right angle on this diagram, justifying your answer.

Angle AFE is 90 degrees because the angle on the circumference of a semicircle is always a right angle.

Next up, step five, pause here to use the sine ratio to calculate the length of AF to as many decimal places as you can.

We know that the diameter of the circle is also the hypotenuse where AE is 18 centimetres.

Therefore, using the sine ratio we have AF at approximately 10.

32438 centimetres.

Step six, do something similar.

Pause here to use the cosine ratio to calculate the length of FE.

Using the cosine ratio, FE is approximately 14.

74474 centimetres.

And finally onto the last step, step seven.

Pause here to use any relevant lengths to find the area of the triangle AEF.

Using the two perpendicular lengths AF and FE, we have 76.

1 centimetre squared as the area of that triangle.

Lovely work so far, onto the practise task.

For question one, split this quadrilateral into two isosceles triangles.

Then find the area of the two triangles in order to find the area of the entire quadrilateral.

And for question two, find the area of triangle FEH by first finding other angles, using circle theorems and identifying similar shapes.

Pause here for these two questions.

Answer question three by first finding the angle KMN, find the area of quadrilateral, JKMN.

Justify your answer with circle theorems and angle facts at each stage.

Pause here for question three.

Onto the answers for question one.

Angle CAB is 53 degrees.

This is because the angle at the centre is twice the angle at any point on the circumference.

Now let's split this quadrilateral into two isosceles triangles.

Angle OAC is 13 degrees because the two base angles in an isosceles triangle are always equal.

The remainder of the 53 degree angle is 40 degrees and once more because base angles in a isosceles triangle are equal, angle ABO is also 40 degrees.

We can split triangle AOB into two congruent right angled triangles.

Let's define P as the midpoint of the line segment AB, the side of that triangle.

The area of triangle OAB can be found by finding the length OP using the sine ratio.

And the length PB using the cosine ratio.

The area of the whole triangle is approximately 70.

91 centimetres squared.

We can then do something very similar with triangle AOC.

Let's define point Q as the midpoint of AC.

Length OQ is found using the sine ratio and length QC is found using the cosine ratio.

The area of this whole triangle is approximately 31.

56 centimetres squared.

Adding these two triangular areas together gives the area of the whole quadrilateral at 102 centimetres squared.

Right onto question two.

Both angles, FEG and GHE are 16 degrees because of the alternate segment theorem.

Angle FEH is 90 degrees because the tangent at point E is perpendicular to a radius or in this case a diameter HE at the point E.

A radius is seven centimetres long and so the diameter HE is 14 centimetres long and FE can be found using the tangent ratio as FE is opposite angle, GHE.

The area of the triangle is 28.

1 centimetres squared.

Onto question three part A.

Angles LKM and KJM are both 50 degrees because of the alternate segment theorem and again, both PNJ and NMJ are 75 degrees because of the same theorem.

The length JM is 70 centimetres as it is a diameter of a circle with radius 35 centimetres.

We have both angles JKM and MMJ at 90 degrees because the angle on the circumference of a semicircle is always a right angle.

Angle KMJ is 40 degrees as it is the final interior angle of a triangle.

Therefore, angle KMN is the sum of 40 degrees and 75 degrees at 115 degrees.

And onto part B.

Let's look at triangle MNJ first.

It's area is 612.

5 centimetres squared.

The area of triangle MKJ is approximately 1,206 centimetres squared.

Therefore the area of quadrilateral JKMN is 1,820 centimetres squared.

Let's now have a look at shapes on circles where angles are given in terms of ratios of each other rather than angles being given explicitly.

Let's have a look.

We can use circle theorems to help us calculate the exact size of angles even if we are given only ratios to show the relationship between angles rather than the actual size of the angles themselves.

So for example, let's find the angle CAB.

To start, let's see if we can find any explicit numerical angles using a circle theorem.

In this case we have angle ACB at 90 degrees because the angle on the circumference of a semicircle is always a right angle.

We are given that CAB and CBA are angles in the ratio two to one.

Therefore, we can consider these two angles algebraically where angle CAB is two X degrees because it corresponds with the two part of the ratio and angle CBA at one X degrees because it corresponds with the one part of the ratio.

We do this for some unknown value of X, which we want to find.

Interior angles in a triangle sum to 180 degrees, therefore two X for the angle CAB, plus X for the angle CBA, plus 90 for the angle ACB equals 180.

This gives us a simple linear equation to solve.

Now that we have X equals 30, we substitute this to get the angles CAB at 60 degrees and CBA at 30 degrees.

Let's have a look at this diagram.

Pause here to identify which of these circle theorems can define a relationship between the angles DCB and DAB.

These two angles are in the same segment because they're both angles on the circumference that have been subtended by the arc DB.

Knowing this relationship and the fact that angles ADE and ECB are in the ratio two to three.

Pause here to identify which of these equations is true for this diagram.

We first label the two angles in the ratio as three X and two X for some unknown value of X.

We also have two pairs of these angles because each pair of angles are in the same segment.

The three interior angles of the triangle, CEB are three X degrees, two X degrees, and 100 degrees, which all sum to 180 degrees giving us this equation.

Pause here once more to find the size of the angle ABC.

Solving the equation from the previous question gives us X equals 16, meaning the angle ABC is twice this angle, two X at 32 degrees.

We can also write down a ratio between different angles using circle theorems, even if we're not given the actual size of many of the angles.

This is because we can use relationships between different angles to form that ratio.

For example, let's find the ratio for these four angles.

When given a question as broad as this, it's best to just label as many angles as we can using any circle theorems or angle facts that are appropriate to that diagram.

Do not worry if the angle you find the size of isn't one of the four in this question.

Chances are it will still be helpful in finding other angles that are part of the question.

Right, so we have angle ABC at the circumference of the circle and angle AOC at the centre, both subtended by the arc AC.

We know that angle AOC is double 72 degrees at 144 degrees.

Furthermore, both angles OAD and OCD on 90 degrees because a radius and tangent meet at the circumference at a right angle.

And also angle ADC equals 36 degrees because we know the other three interior angles of the quadrilateral, AOCD and the interior angles of any quadrilateral, sum to 360 degrees.

We now know all four angles in this question.

These are the angles which can be simplified to the ratio five to two to five to eight.

Okay, here's a really quick check.

Pause here to identify the ratio X to Y in its simplest form and can you justify your answer with a circle theorem.

Angles X and Y are equal because angles in the same segment are equal, therefore the ratio is one to one.

And for the same diagram, pause here to explain whether it is possible or not to find the ratio X to Z with the given information.

It is not possible because these two angles are not related by any circle theorem or angle fact.

Great stuff onto the practise for this cycle.

For question one, find the ratio of the angles, UOR and ORU given that the angle UOT is 84 degrees.

And for question two, given that the angles MOL and NJK are in the ratio two to three, label as many angles in this diagram as you can.

Pause here for these two questions.

Onto question three, find the ratio of these three interior angles of the triangle ABC.

And for question four we have three interior angles of the triangle GEH.

Find the size of all three interior angles and label each angle in the correct location on the diagram.

Pause again for these two questions.

Great effort, everyone.

Here are the answers.

For question one angle UOR is 96 degrees as angles UOR and UOT lie on a straight line because triangle UOR is an isosceles triangle angles ORU and OUR are equal.

Therefore angle ORU is 42 degrees.

Angles UOR and ORU are therefore in the ratio 16 to seven after being simplified.

Okay, onto question two.

Here are the angles MOL and MJK.

Angles MOL and KON are vertically opposite angles and JMO is 90 degrees because the radius and tangent to a circle are perpendicular at the point that they meet.

We can then construct this equation for triangle JNO two X plus three X plus 90 equals 180.

Solving this gives us X equals 18.

We can then substitute X equals 18 to get these angles and using other angle facts, we get these angles as well.

Next up, question three.

Angle ABC is 81 degrees due to the alternate segment theorem.

Angle BAD is 90 degrees because the tangent at any point on a circle is perpendicular to the diameter at that point.

Therefore, angle BAC is the difference between the right angle and the 81 degree angle at nine degrees.

Angle ACB equals 90 degrees because the angle at the circumference of a semicircle is a right angle.

Taking all of this into account, we have the ratio one to 10 to nine when simplified.

And lastly question four.

Angle EGH is 90 degrees and angle GHE is the smallest angle of the triangle because it is opposite the shortest side of the triangle.

Therefore angle GHE has the smallest angle at two X and angle EGH has the largest at five X as no angle can be greater than 90 degrees in a right angled triangle.

Knowing that angle EGH is 90 degrees, we can identify that X is 18, giving these two angles, GHE at 36 degrees and GEH at 54 degrees.

And for our last cycle, let's go one step further than ratios with our circle theorems. Let's see how we approach questions where basically everything is given as an algebraic expression.

For one last time, let's have a look.

If the size of an angle is given as an algebraic expression, then we can use circle theorems and angle facts to write the size of a different angle in terms of the original one, again, as an algebraic expression.

For example, let's try and write the angle BAC in terms of the angle X.

Sometimes this is tricky or even impossible with the line segments given on a diagram.

Drawing on one extra line segment can make some questions much, much easier to solve.

So pause here to identify which line segment could be drawn to help with this question and explain why it helps.

Drawing either the line segment CD or AD can be added to help us.

Let's focus on line segment CD for the moment.

As with all questions like this, let's try and find as many relevant circle theorems and angle facts as possible to identify as many different relationships between angles as possible.

When we've found a few, then we can use these relationships in order to answer the question.

If line segment CD is added, then we have a right angled triangle, BCD, where BD is both the hypotenuse of the triangle and a diameter of the circle.

Furthermore, we have a second triangle, ABC with a shared side BC that subtends angle BAC, in the same line segment as angle BDC from the other triangle.

Since these two angles are in the same segment and both subtended by the chord BC, angles BDC and BAC are equal to each other.

Let's also look at more angles within one of those triangles, starting with triangle BCD.

Angle BCD is 90 degrees since it is on the circumference of a semicircle.

This means that the other two interior angles of this triangle have this following relationship.

The two angles must sum to 90 degrees, which gives X degrees plus the angle CDB equals 90 degrees.

This means that the angle CDB can be written as the expression 90 take away X degrees.

Right, let's bring all of our calculations together.

We know that angles CDB and BAC are equal, therefore, they are both equal to the expression 90 take away X degrees.

The expression 90 take away X degrees is the answer to this question as an algebraic expression.

So pause here to explain how instead we could have come to the same conclusion if we'd focused on drawing chord AD instead of chord CD.

Angles CBD and CAD are angles in the same segment as they are both angles subtended by arc CD.

And since angle BAD is a right angle at the circumference of a semicircle, we can find angle BAC as the difference between the 90 degree angle and the angle X.

Right, let's tackle different angles in this diagram one by one.

First up, pause here to find the angle DCB in terms of X.

Angle DCB has the expression 180 takeaway three X degrees.

Next up, pause here to find the angle EDC, again, in terms of X.

We have the triangle EDC and so angle EDC is 180 degrees take away X degrees for angle DEC.

Take away this expression 180 take away three X degrees for angle DCE.

This then simplifies all the way down to two X degrees.

And lastly, pause here to find angle ABE once more in terms of X.

Angle ABC is opposite the angle two X degrees in a cyclic quadrilateral.

Therefore angle ABE itself is also two x degrees because the angles ABC, and ABE lie about a point, B, on a straight line.

Amazing perseverance everyone on these really challenging problems. Here's the final practise task.

For both questions one and two, find the size of an angle as an algebraic expression in terms of either alpha or beta.

Remember to label as many angles as you can with algebraic expressions to help you.

Pause now for these two questions.

And finally question three, find an algebraic expression for the size of angle COB in two different ways, once using a circle theorem and another using a different angle fact.

You may have to draw on an extra line segment to help you.

Pause here for this final question.

Great stuff, let's get ready for the answers.

For question one we have a 90 degree angle where a tangent and a radius meet at the angle ECO.

Therefore angle DCO has the expression 90 takeaway alpha degrees.

Both angles CDO and DCO are equal because they are the base angles in an isosceles triangle.

Angle COD is the remaining interior angle of an isosceles triangle at two alpha degrees.

An alternate method can involve you identifying a random point on the circumference of the circle in a major segment created by the cord CD.

Angle CPD is alpha degrees due to the alternate segment theorem and because angle COD is at the centre and subtended by the same cord CD, it is double alpha at two alpha degrees.

Onto question two, we have angles KNJ and KLN that are both equal to each other both at beta degrees because of the alternate segment theorem and the chord KN.

The angle KNL is the third interior angle of a triangle giving an angle of 170 take away four beta degrees.

Angle LMK is opposite angle KNL in a cyclic quadrilateral.

Therefore its expression is 10 plus four beta degrees.

Triangle LNK is an isosceles triangle where both angles MLK and MKL are equal to each other.

Therefore, angle MLK is 85 degrees take away two beta degrees.

And onto the final question three.

For part A, we can draw on line segment OA.

This splits the quadrilateral into two isosceles triangles.

Because these are both isosceles triangles, angle OAC equals X degrees and angle OAB equals two X degrees.

Therefore the combined angle CAB is three X degrees.

Our circle theorem is the angle at the centre is twice the angle at any point on the circumference of a circle.

Therefore angle COB is the angle at the centre of the circle at six X degrees.

For part B, we need to first find the non-base angle in each of the two isosceles triangles.

Angle AOC is 180 degrees, take away two lots of X degrees and angle AOB is 180 degrees take away two lots of two X degrees or 180 degrees take away four X degrees.

Therefore, the reflex angle COB is the sum of the two angles at 360 take away six X degrees.

Our angle fact is that angles about a point sum to 360 degrees.

Therefore the angle COB itself is six X degrees.

The same answer as we got using the circle theorem.

And for part C angle DCO equals 90 degrees because C is the point where a radius and a tangent meet and they must meet at a right angle.

Therefore angle ACO equals X degrees and X degrees equals 90 degrees, take away the 70 degrees giving us 20 degrees, meaning the X equals 20.

If X equals 20, then the angle COB is six X degrees or 120 degrees.

Blimey, that was a really intense lesson and a super well done for persevering through it and weaving circle theorems into questions involving Pythagoras' theorem and trigonometry in order to find lengths and areas of shapes, and also setting up and solving equations on circle diagrams that involve ratios of angles.

And rather than numerical angles, we've looked at how to write one angle in terms of another, where both angles are given as our algebraic expressions.

Once more, thank you all so much for your effort in this lesson and during every single one of the lessons in the circle theorems unit.

I have been your teacher, Mr. Gratton, and you have all been pretty spectacular.

Until our next maths lesson together everyone, take care and have an amazing rest of your day.

Goodbye.