Hello everyone and welcome to Circle Theorems Part 4. In this tutorial we will focus on hints and tips to tackle exam questions using our knowledge on these circle theorems. Each question will use more than one circle theorem and remember in mathematics there's usually more than one way to get the right answer. So let's have a look at a question. This question states A, B, C, D are points on the circumference of a circle with a centre O. We know AC is a diameter of the circle and AC and BD intersect at point E.
The angle CAB is 25 degrees and the angle DEC is 100 degrees. We're asked to work out the angle D to A to C and we must show all our working out. See if you can give it a go and press pause if you need. It's important to extract as much information as possible to determine which circle theorems we can use.
Here you can read and see we do not have any tangents. No angles at the center, no isosceles triangles from radii, and no cyclic quadrilateral. Therefore, we have two remaining circle theorems that we could potentially use in this question.
Now let's move our circle theorems here. Reading the question, you might notice we have a diameter. Therefore, angle ABC is 90 degrees. This is because angles subtended at the circumference by a semicircle is always 90 degrees. From here, we can work out angle ACB by simply doing 180, subtract our 90, subtract our 25, giving angle ACB to be 65 degrees.
This is because angles in a triangle sum to 180 degrees. Now let's see if we can spot our other circle theorem. You might be able to notice angle ACB is exactly the same as angle ADB.
This is because angles subtended by the same arc are equal. Therefore, we know angle ADB is 65 degrees. Now, let's see if we can use this 100 degrees. We can work out angle AED.
180 subtract our 100 gives us an angle of 80 degrees. This is because angles on a straight line sum to 180 degrees. Now, we can finally work out angle DAC.
180 subtract our 80 subtract our 65 gives us a final answer of 35 degrees. This is because angles in a triangle sum to to 180 degrees. Moving on to our next question.
This question states B, D, E and F lie in a circle. A, B, C is the tangent to the circle at point B and we're asked to find angle A, B, D. It's important to give a reason for each stage of our working. Why not give it a go and press pause if you need. So let's have a look at our circle theorems. We need to extract as much information as possible from our question and diagram to determine which circle theorems we can use.
Here you can read and see that we do not have any radii, we do not have the diameter, we do not have any semicircles, nor do we have any angles subtended by the same arc. This leaves us with the possibility of using the alternate segment theorem and angles in a cyclic quadrilateral sum to 180 degrees. So let's put these circle theorems to the side.
Now looking at the question you might be able to spot that we have a cyclic quadrilateral. So we can work out angle D B F to be 180 subtract 100 which gives us an angle of 80 degrees. This is because Opposite angles in a cyclic quadrilateral sum to 180 degrees.
Now from here, we can calculate angle BFD. 180 subtract our 80 subtract our 40 gives us an angle of 60 degrees. This is because all the angles in a triangle sum to 180 degrees. Now we can have a look at our last circle theorem, and we might be able to spot the alternate segment theorem. Here you might be able to see angle ABD must be 60 degrees because using the alternate segment theorem it's equal to angle BFD.
Let's move on to the next question here it states B, C, D are points on the circumference of a circle with center O we know ABE and ADF are tangents to the circle We're given angle DAB to be 40 degrees and angle CBE to be 75 degrees. We're asked to calculate angle ODC and we must give reasons for our answer. Why not give it a go and press pause if you need. So let's have a look at our circle theorems and extract as much information from our question and diagram to determine which circle theorems we can use.
Here you can see we have no diameter, we have no cyclic quadrilateral, we have no angles subtended by the same arc, we have no alternate segment theorem and we have no triangles made from radii. Therefore, it leaves us with these three potential circle theorems. So let's move them to the side and see if we can tackle our question. It's clear we can use tangents meet the radius at 90 degrees.
Therefore, angle ADO is 90 degrees and ABO is 90 degrees. This is because we know our tangents meet the radius at 90 degrees. Now from here, we can work out angle DOB. 360 subtract our 90, subtract our 90, subtract our 40. Gives us an angle of 140 degrees.
This is because angles in a quadrilateral sum to 360 degrees. Now let's see if we can use another circle theorem. Well, we know angles at the center are twice that of the circumference.
So therefore we can work out angle DCB to be 70 degrees. because 140 divided by 2 is 70. Let's see if we can calculate the angle OBC. Well, if we know this is 90 degrees, 90 subtract our 75 gives us an angle of 15 degrees.
This is because we know tangents meet the radius at 90 degrees. Now, let's see if we can work out our reflex angle of angle DOB. 360 subtract our 140 gives us an angle of 220. Now we can find angle ODC.
360 subtract our 220 subtract our 70 subtract our 15 gives us a final answer of 55 degrees. This is because angles in a quadrilateral will sum to 360 degrees. Now let's move on to our last exam question.
It states A, B, C, D are points on the circumference of a circle with a centre O. We know F, D, E is the tangent of the circle. We're asked to show that Y subtract X is equal to 90 degrees.
And we must give a reason for each stage of our working out. Why don't you give it a go and press pause if you need. Here it's important to extract as much information as possible from the question to identify which circle theorems we can use.
Here you can see we do not have a diameter, no point outside of the circle, no angle subtended by the same arc. Therefore, that leaves us with five potential circle theorems, indicating there is more than one way to show this answer. For now, I'm going to show you one way in which we can achieve y minus x equals 90 degrees. Firstly, let's move our circle theorems to the side. Now, looking at our diagram, you can see we have a radius.
It looks a little incomplete, so let's add a second radius. From our diagram, you might be able to see that we have a cyclic quadrilateral. Therefore, we can work out angle B, C. d.
This is 180 subtract y. This is because opposite angles in our cyclic quadrilateral will always sum to 180 degrees. Before we added the length OB as our radius, you might be able to see OB is equal to OD.
And you can see our arrowhead shape. From here, recognizing our circle theorem, we know the angle at the center is twice that at the circumference. So therefore, 2 multiplied by 100 minus y gives us angle BOD to be 360 minus 2y.
This is because the angle at the center is twice the angle at the circumference. Now you can see we have our isosceles triangle. So therefore, we know angle OBD must be x. So we can work out the value of x from our Cecily's triangle. Here you can see I've written an equation.
180 degrees is equal to x plus x plus our 360 subtract 2y. From here, I've simplified. Now I'm moving the y term and the x term onto the right-hand side, and I've subtracted my 180. Here I now have...
180 is equal to 2y subtract 2x. Dividing everything by 2, I end up with 90 equals y minus x, which is what the question asked us to show. There's more than one way to show y minus x equals 90. So see if you can find it using an alternative method, ensuring to show each stage of your working out. So, here is a summary of...
all the circle theorems you need to memorize and our key angle fact. Remember, when tackling exam questions, always pull out key words from the question to isolate which possible circle theorem you might be using. If you like this video, please give us a thumbs up.
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