hey there what's up guys so to continue with our Circle geometry theorems here I've prepared six examples for us to quickly go through and for each example I'll give you like a second to pause the video and attempt it on your own and then press play when you're ready to see the solution now on the screen right now you'll see all the theorems that you'll need to know for this video so first we have our Center theorems angle at the center two times angular circumference you have your line from Center perpendicular to chord bisex code you also have your angle in a semicircle and radii which creates these angles opposite equal sides then moving on to our cyclic quad theorems here you'll see opposite angles add up to 180 angles in the same segment exterior angle of a cyclic quad and also this theorem that's not necessarily linked with cyclicons but I just grouped it with this punch so it's equal chords subtained equal angles and lastly we have our tangent theorems here we have the tan code theorem then two tangents from the same point are equal in length also radius perpendicular to tangent okay so these are the theorems you'll need to know to answer the questions let's start with our first example you can ignore the fine numbering that you'll see on the side for example the 3.3 in this one I've extracted these questions from the answer series textbook so the numbering will sort of jump around as I've taken these examples from different sections in the textbook okay so for example number one we have a circle going through the points a b and c which also forms a triangle then point O is the center of the circle angle a is 40 degrees and angle is 15 degrees you need to determine with reasons the size for angle X Y and Z okay so pause the video now if you'd like to give it a try and press play when you're ready to watch the solution okay if you're ready let's see the solutions so angle X is at the center of the circle now we know that angles at the center of a circle will be twice the size of the angle subtended at the circumference of the circle so therefore angle X will be equal to twice the size of angle a and since angle a is 40 degrees angle X will be 80 degrees okay that simple to calculate why we first need to understand that OB and OC are radi now this creates an isosceles triangle okay so the angle by C would also be equal to Y you have these base angles B and C um while it's OBC and OCB both of them are equal to Y and they're now also using the theorem that interior angles of a triangle will sum to 180. we can then work out y as 50 degrees you simply say 180 minus the 80 degrees divided by 2 and this gives us 50 degrees for angle y all that so now to find Z in triangle ABC we know that the sum of the angles of that triangle must give us 180 so by minusing the 40 degrees the 15 degrees and the two angles of 50 degrees will get the size of angle Z and so the answer to that is 25 degrees okay moving on to example number two so again pause the video if you'd like to give this a try on your own and press play when you're ready to see the solution okay ABCD is a cyclic chord opposite angles of a cyclic quad should sum to 180 degrees so therefore lowercase a would be a hundred degrees and lowercase b would be 60 degrees opposite angles of a cyclic quad example number three this one is a lot more challenging so definitely pause the video here to give it a try on your own okay so here you have the circle going through points l m in and K and you have this tangent PMT right they also give us the size of angle LMN which is 35 degrees and O is the center of the circle we need to find the size to the angles a to e in alphabetic order so we'll start with angle a angle a is the angle between the tangent and a chord we have our tangent PMT we have the code LM subtended from the code is the 35 degree angle by LMN this means that angle a is also 35 degrees since we have angle between a tangent and a chord equal to the angle subtended from the chord also known as the tan code theorem now to find angle B m n is the diameter of the circle we know that diameters always subtained 90 degree angles at the circumference of the circle so angle B is 90 degrees the name of this theorem is angle in a semicircle then if you look at angle C that is also 90 degrees notice PMT is a tangent and om is a radius and those two lines form angle C so where radius and the tangent meter always perpendicular therefore angle C is 90 degrees now to find angle D is pretty straightforward you have angles on a straight line summing to 180 degrees you can also think of d as angles in a triangle sum into 180 degrees either way you get angle d as 55 degrees now to find e you had to see the cyclic quad LMN and K opposite angles of a cyclic quad adds up to 180 degrees so here we can see that angle e is 125 degrees and that's it for that example example number four here we have a circle passing through x y l and m with the center o and o l is parallel to m l and angle X is 40 degrees we need to find the size for angle o1 Y2 and O2 with reasons pause the video now if you'd like to try this on your own but if you're just here to enjoy the show you can keep watching so angle 01 is the angle at the center of the circle angle at the center of a circle is equal twice the angle at the circumference can you see that angle X is the angle at the circumference so therefore angle o1 will be twice the size of angle X which makes it 80 degrees to find Y2 again you can see the same theme playing out here o y and o l are radii of the circle and so we have the same scenario like in the first one where we have this isosceles triangle so angle Y2 and angle Oly will both be equal to 50 degrees okay sum of angles in a triangle what are the angles opposite equal sides the isosceles triangle theorem okay then to find O2 we need to now use our parallel lines notice o y is parallel to L in so you can form these alternate angles o1 is alternate to olm so angle olm is 80 degrees again we can take into account that ol and om are radar of the circle so they are equal to each other and you have this isosceles triangle again so angle m is also 80 degrees which therefore makes angle O2 20 degrees since that's the sum of angles in a triangle okay pretty simple stuff so far I hope you're getting used to this let's move on to the first example so here we have a circle passing through points a b c and d then we have this line RDS which is the tangent you can see it at the bottom there and angle D3 is given as 10 degrees Point O is the center of the circle calculate with reasons the sizes of angle B2 o1 a c B1 and D1 pause the video now if you'd like to give the problem a try okay I hope you're getting more comfortable with this now OD and OB are radi to the circle that makes angle B2 also 10 degrees angles opposite equal sides then you have angle 01 which equals 160 degrees this is the sum of angles in a triangle angle a with therefore be 80 degrees since we have angle at the center equals two times the angle at the circumference and angle C is going to be 100 degrees since we have opposite angles of a cyclic quad now it's also given that BC is equal to DC therefore we have another isosceles triangle so angle B1 and angle D2 will both be equal to 40 degrees sum of angles in a triangle angles opposite equal sides and lastly angle D1 which is the angle between the tangent and the chord and we know that it's equal to the angle subtended from that code so angle D1 is equal to B1 both of them are 40 degrees so even though this question looked quite complicated it was pretty simple to do what we should also keep in mind when doing these questions is that there are definite ways to finding the answers you might have arrived to the answer in a completely different way but if your reasoning is still sound and correct then you will still get the answers even though you've chosen different paths okay let's check our last example so for our last example again we have a circle with Center o and points a t c b all lying on the circumference of the circle line sdu is a tangent to The Circle at Point T angle ATC is 105 degrees and angle CTU is 40 degrees also BC is equal to TC and that's going to be helpful for the first question now you need to calculate the size of angle A1 o1 and angle B2 pause the video now if you'd like to give it a try and press play when you're ready to see the solution okay so to find o1 this was kind of tricky but I did hint that you'd have to use the equal codes first you had to see that the angle C to you 40 degrees is the angle between a tangent and a chord and that angle is equal to the angle subtended by the code so angle A2 is also 40 degrees now chord TC subtains angle A2 and chord BC also subtains angle A1 since they are equal equal chords will subtained equal angles so angle A1 is also 40 degrees then the rest becomes easy if angle A1 is 40 degrees then angle o1 will be 80 degrees you've seen this countless times angle at the center equals 2 times the angle at the circumference but to find angle B2 we actually first had to find the angle B1 so what I did is I said since OB is equal to OC they both are a die you have this isosceles triangle so angle B1 and C1 will both be 50 degrees angles opposite equal sides then if I have a look inside the cyclic quad atcb B and T are opposite angles in the cyclic quad so the whole angle B plus the angle by T would be equal to 180 degrees now since angle ATC is 105 degrees the whole angle B should be 75 degrees and this makes angle B2 25 degrees since we already have B1 as 50 degrees I hope that makes sense okay this is the last example I hope you guys enjoyed this video maybe learn something today and had fun solving these problems I'll see you in the next video