I dont mean to be so off topic but does someone know of a way to log back into an instagram account..? I somehow lost the login password. I appreciate any tips you can offer me.
@@howardtalon2255 umm it's not an right place to ask that question 😅 But you can reach out to them by going to the Instagrams help section and than follow thier advice to get the account back ig....
I've been struggling with angles in circles for about 2 months now, I've watched videos upon videos on the topic and even bought textbooks, but nothing has helped me as much as this video did. You're a saint, so happy a found this!
I have a 15 year old daughter about to write her mock exams and she found it hard to get circle theorem, now she's learning and begining to get it. Mathematics is not a problem for her but this one hit her hard. Thanks for the explanation. God bless
This video was so helpful ! I loved the way you explained this so easily. I have an exam in a few hours and I finally understand circle theorems. Thank you !
THANK YOU SO MUCH! I always had trouble with circle properties and this helped me so much to gain confidence and learn all the properties! Such a great video!
Wow, amazing, fabulous teacher, thanks so much for helping me out and I wish you all the best in your teaching on social media, and may you kindly continue your great work and help to everyone... May God Bless you Sir🙏
I can now solve questions on circle theorem without any problems after watching this video. I am kindly requesting if you can make a video to show how to answer questions on variation as well.
Thank you so much sir. I have been finding this concept difficult since grade 10 and I have finally understood it . Thank you so much. Now I can confidently solve any question on this.
i give thank to GOD for meeting this great mathematics lecturer. Sir thanks you for expanding my understand on circle theorem. In all the topics of maths circle theorem is my problem but by meeting it give fit to challenge circle theorem
you reaaaally saved meee.... its the day before the half yearly examinations and I was so stuck with these stuff but you really saved me! Thanks! Now I can give a better exam!
I have a question about 39:37 why you couldn't do the same for the opposite angles which would mean that angle k+the unknown angle=90 degrees. meaning that the radius from ''o'' to 34 degrees and the unknown angle is equal to line ''o-k'' so 34+k=90 meaning K=56 and then since there are 2 k's. angle at ''l'' would be equal to=68 degrees (This was just a thought because i found it easier) Lastly at - 44:48 - is angle ''s'' not (r+70) thanks
I would recommend watching Kerwin Springer video on introductory to circle theorems, and draw all 8 theorems. Then do the practice exercises here. It will be much simpler. Also for some of the theories just use the alternate segment theory which takes no working you just have to look and put the answer.
Good job...but one slight error. At 44:55 the third figure isn't a cyclic quadrilateral inscribed in a circle. The figure inscribed is a 5-sided one and therefore cannot be used to determine angle 46 at the top. The angle of 134 does not belong to the cyclic quad.. To derive the 46 degree one would have to extend 2 radii from the centre to meet the chord above angle 134 degrees....and then find the reflex angle at the center which would be 2 x 134 = 268.....we then use the other angle at the center at the angle being 92. to derive the 46.....
Wooow I really enjoyed this video. It was fun as if were matching pieces of a puzzle. You could hear me shouting "that angle,that angle, yes that one "🤣 oh my . ...i now understand this topic you got yourself a happy subscriber .Stay blessed teacher 🙏
20:29 two diameters are intersecting it would create a 90 degrees at its intersection, shouldnt K be 36 and its adjacent angle is 54 so that if it was added it would form 90 degrees?
I appreciate your question. However, we are dealing with a semicircle trangle at the circle which make angle 90 at the circumference. So k will naturally be 54 degree as calculated
Thanks alot bro,this helps,I have a test tomorrow for 30marks. This helps me to inderstand some of the theroms and how to solve them. Thankyou brother.
Thank you so much for the asking, I just realized i didn't add that to the original video. Anyways, the reasoning for the angle APB is this : ANGLES IN A QUADRILATERAL ADD UP TO 36O
Could the one at 53:38 be solved by the rule that the angle at the center is twice the angle at the circumference or would that only work if the angle was in the triangle (in a different triangle since the rule does not apply to the same triangle)?
Thanks for your question. I presume you may be asking how the value of n is solved. To use the circle theorem, that is, angle at centre = 2 x at circumference wouldn't suffice to give the solution for n. You will have to consider and use the circle theorem which state that the angle between tangent and radius of circle is right angle to solve it
@@ICOMATHS Thanks for the answer. I fully understand the theorems (thanks to your previous circle thereom video and practice in this one). I just was bringing it up because the value of n (81) was half the value of O (162), and n was at the circumference.
Congratulations teacher. You are actually a good instructor. Can I know how you got used of the pencil you are writing with I wish to use it as well in my teaching. I am getting trouble on using it.
Thank you for your inquiry. I use Wacom pen and tablet www.wacom.com/en-us. You get better writing with it when you use it every so often. I think practice make it easier
Thank you so much for the question. it seems so but not quite correct. Angles AOC and ABC are not on the same segment. Infact, angle AOC is not on a segment at all. it just subtends an angle at the centre O.
44:07 why did you say 180-134 when it's not a cyclic quad ? A cyclic quad has 4 points on the circumference but in this case you used a Pentagon , why?
Thank you so much for your question. Apologies for replying late. Now i think there are two scenarios playing out here. The first is a cyclic quadriateral wherre the opposite angles are supplementary that is 180 degree. The second is an issoccles triangle where the base angles should be the same. In all those two scenarios i expalined how the various angles can be calculated following the rules
Hello, I have a concern. For number 9) where you are asked to find r and s I understand how you got the 23 and the 70 but for s wouldn’t you add the 70 and the 23 seen as angle s is from the quadrilateral and then there is a little part in the triangle?
just got to this question myself and had exactly the same issue - i'll let you know if i figure it out but i'm kinda shite at circles so don't get ur hopes up x
at the timeline cited, i dont think i used the cyclic quadrilateral theorem. perpharps you may cite the exact timeline for me to see exactly where you are referring to. Anyway thank you so much for your feedback. They are appreciated
31:06 angle subtented on a semicircle =90 deg. therefore 90+2i=180 2i=180-90 -----> 2i=90 --->i=90/2 =45deg. sir I'm not sure. is my answer correct at this point....
There is therum. Angle subtended by an arc of circle at its centre is double the angle subtended by the Sam arc at remaking circumference. X = 110 Degrees and angle at C = 70 degrees
ok thanks for your question. You see, why i subtracted 36 from 180 and then halfed the result was because the triangle is an isoccesles triangle. The base angles of this type of triangle should be the same.
Hello, I don’t quite understand why you divided the 304 Degrees by 2, it would be much appreciated if you could reply to this comment and explain, thanks. Additionally I would like to point out that you could also use the circle theorem that states that the angle subtended(created) at the center by the angle at the circumference in the same segment I.e on the same chord is twice the angle at the circumference. This simply means that you could’ve divided 56 degrees by two and you would arrive at the same 28 degrees. Thanks for reading my comment and thank you for the vid.
Ooh my gosh u are the best I have exams coming up and I couldn’t understand a thing I was soo lucky to find this before my exam thanks I now get it
Thanks so much for this
You are the best
I have exam tmr...this guy is a lifesaver😂
Mi too
I wish i had you as my teacher back in high school days. You make maths easy to understand
Thanks so much for this
I dont mean to be so off topic but does someone know of a way to log back into an instagram account..?
I somehow lost the login password. I appreciate any tips you can offer me.
@@howardtalon2255 umm it's not an right place to ask that question 😅
But you can reach out to them by going to the Instagrams help section and than follow thier advice to get the account back ig....
I've been struggling with angles in circles for about 2 months now, I've watched videos upon videos on the topic and even bought textbooks, but nothing has helped me as much as this video did. You're a saint, so happy a found this!
Thank you so much
You need to be awarded for this explanation 💯🙌
Thank you so much for the compliment
I have a 15 year old daughter about to write her mock exams and she found it hard to get circle theorem, now she's learning and begining to get it. Mathematics is not a problem for her but this one hit her hard. Thanks for the explanation. God bless
Thank you so much 💓 for the feedback
This video was so helpful ! I loved the way you explained this so easily. I have an exam in a few hours and I finally understand circle theorems. Thank you !
wow. so happy you found the video helpful
Can get the pdf pl
no de lie to am
Taking this lesson from Nairobi, Kenya. I love the way you are doing it.
Thank you so much to hear you find this video very useful
Thank you so much to hear you find this video very useful
THANK YOU SO MUCH! I always had trouble with circle properties and this helped me so much to gain confidence and learn all the properties! Such a great video!
Thank you so much for letting us know how this video has helped you
Wow, amazing, fabulous teacher, thanks so much for helping me out and I wish you all the best in your teaching on social media, and may you kindly continue your great work and help to everyone... May God Bless you Sir🙏
Thanks so much for sharing your thoughts. it is highly appreiated
@@ICOMATHS ua-cam.com/video/GphwjWJ5WwI/v-deo.html
I can now solve questions on circle theorem without any problems after watching this video. I am kindly requesting if you can make a video to show how to answer questions on variation as well.
Thankyou so much, solving theoroms have always been a struggle for me. This video helped a lot!
Thank you so much for letting us know. I appreciate all
You are the best. Before I don't understand the circle theorem and now I understand the circle theorem. Thank you so much
Wow. This is good news
Man you are a legend all thanks to you i have a chance for passing my midterm !!
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God bless you richly. Something I have not understood for days, I have gotten in an hour. I am very glad. Bravo!!! ❤
Thanks so much for letting me know
Your lessons helped me over 4 years, thanks alot
I am happy to hear this helped you. God bless
You know what you teach and teach what you know sir . God richly bless you 🙏
thanks so so much
watched the entire video , found it helpful and thank you very much , god bless you
Wow. That was inspiring
You are a wonderful teacher. You explanations right on the point. I will never fail Maths with your help. Thank you so much
Thank you so much 💓
Thank you so much sir. I have been finding this concept difficult since grade 10 and I have finally understood it . Thank you so much. Now I can confidently solve any question on this.
Thanks for finding this video super helpful. I am touched by it
i give thank to GOD for meeting this great mathematics lecturer. Sir thanks you for expanding my understand on circle theorem. In all the topics of maths circle theorem is my problem but by meeting it give fit to challenge circle theorem
Wow , thanks so much for the feedback,
Thank you sir you are literally a lifesaver, I hope that you live a blessed and happy life!
Thank you so much 💓
you reaaaally saved meee.... its the day before the half yearly examinations and I was so stuck with these stuff but you really saved me! Thanks! Now I can give a better exam!
I am so happy to hear you find this video helpful. Wishing you the best in your exams
best math teacher
Thanks so much
wowwwww these practice questions actually helped ALOTTTTT
Wow, thank you!
i am a bit too late, but i found your video extremely enjoyable and very resourceful!!! Thank you so much for the help!!!!
Wow. Thank you so much
i know im late but this really helped alot!!! Im really bad at propertioes of circles so this helped by a ton thank you!!!
Thank you so much. I am so happy you found it helpful
this video is so helpful and i can see some theorems you did apply like exterior angle of a cyclic quad
Thanks alot for your thoughts
Damnnn this vid is quite good for covering questions of the topic!!!
Thanks a lot of this vid ☺️
you are welcome
Superb Explanations, Great Video
Thanks so much for the feedback
Thanks for this explanation. Now I understand better.
Glad it helped!
using this the day before the exam... and i finally get it!
wow
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Great
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Do so more please 🙏
I have a question about 39:37 why you couldn't do the same for the opposite angles which would mean that angle k+the unknown angle=90 degrees. meaning that the radius from ''o'' to 34 degrees and the unknown angle is equal to line ''o-k'' so 34+k=90 meaning K=56 and then since there are 2 k's. angle at ''l'' would be equal to=68 degrees
(This was just a thought because i found it easier)
Lastly at - 44:48 - is angle ''s'' not (r+70)
thanks
Sir grt , u explain so well , I enjoy doing it. It’s like a game.
Thank you so much
Ur such a amazing teacher tbh❤
Thank you so much
THANK YOU SOOO MUCH!! Your video has helped me so much I really appreciate it!!!
Thank you so much. I am so happy you found this video helpful
I would recommend watching Kerwin Springer video on introductory to circle theorems, and draw all 8 theorems. Then do the practice exercises here. It will be much simpler. Also for some of the theories just use the alternate segment theory which takes no working you just have to look and put the answer.
The alternate segment theorem is the angle between the chord and the tangent is equal to the opposite angle inside the triangle.
U are the best teacher bravoooooooo
Thank you so much for the feedback. I am happy you find the video helpful
Thank you so much for this video, it had strengthened my conceptual understanding of circles!!
Wow
U are the best finally understood this topic today❤❤❤
Wow. Thanks alot
I really enjoyed your, how I wish you were my mathematics teacher back before now.
Wow, thank you!
@@ICOMATHS please sir help with proofs of the theorems.
@@alhajiswarraymoiwah6740 Thanks for your question. I hope one day to do this
Extremely helpful, very clear - thank you
Thank you so much
fantastic dear lovely teacher may God protect you
great and appreciate the prayers
Thank you sir...made me understand it really well
Thanks alot for letting me know. I appreciaate it
I'm truly enjoying your lessons...
Thanks so much for letting me. i appreciate you find it useful
This video helped me a lot 🙏. Circle theorem made easy ; thank you 😊
wow
Wow, I really understood everything. Thanks.
Wow, this is so encouraging
Good job...but one slight error. At 44:55 the third figure isn't a cyclic quadrilateral inscribed in a circle. The figure inscribed is a 5-sided one and therefore cannot be used to determine angle 46 at the top. The angle of 134 does not belong to the cyclic quad.. To derive the 46 degree one would have to extend 2 radii from the centre to meet the chord above angle 134 degrees....and then find the reflex angle at the center which would be 2 x 134 = 268.....we then use the other angle at the center at the angle being 92. to derive the 46.....
Thank you so much for the
spot on the minor error. I appreciate your feedback very much.
Thank u sir it’s well understood and appreciated
Thank you so much for letting me kmow how the video helped you
Wooow I really enjoyed this video. It was fun as if were matching pieces of a puzzle. You could hear me shouting "that angle,that angle, yes that one "🤣 oh my . ...i now understand this topic you got yourself a happy subscriber .Stay blessed teacher 🙏
Glad you enjoyed it!
i am really grateful to this video for enlightening me🤟🤟🤟🤟🤟🤟🤟🤟
So happy you find the video helpful
20:29 two diameters are intersecting it would create a 90 degrees at its intersection, shouldnt K be 36 and its adjacent angle is 54 so that if it was added it would form 90 degrees?
I appreciate your question. However, we are dealing with a semicircle trangle at the circle which make angle 90 at the circumference. So k will naturally be 54 degree as calculated
This is really great teaching 😊
Thanks
that was so helpful..thank you so much for this☑
Glad it was helpful!
U are amazing , u help me a lot by this video sir 😃☺☺🙂😃
Glad to hear that
This was more useful than my teacher
Thank you so much for the feedback
this video is so helpful. The best. Thank you
Glad it was helpful!
This was really helpful
Thanks so much
Thank you so much
Thanks alot bro,this helps,I have a test tomorrow for 30marks. This helps me to inderstand some of the theroms and how to solve them. Thankyou brother.
Thank you so much for this feedback. I am so happy you found this viodeo helpful. Wishing you the best in your test
at 15:02 what is the reasoning for question b (workout the size of angle APB)
Thank you so much for the asking, I just realized i didn't add that to the original video. Anyways, the reasoning for the angle APB is this : ANGLES IN A QUADRILATERAL ADD UP TO 36O
Could the one at 53:38 be solved by the rule that the angle at the center is twice the angle at the circumference or would that only work if the angle was in the triangle (in a different triangle since the rule does not apply to the same triangle)?
Thanks for your question. I presume you may be asking how the value of n is solved. To use the circle theorem, that is, angle at centre = 2 x at circumference wouldn't suffice to give the solution for n. You will have to consider and use the circle theorem which state that the angle between tangent and radius of circle is right angle to solve it
@@ICOMATHS Thanks for the answer. I fully understand the theorems (thanks to your previous circle thereom video and practice in this one). I just was bringing it up because the value of n (81) was half the value of O (162), and n was at the circumference.
@@Speedy_Ethan thanks. your observation well noted
Sir I am not understand alternative angles are equal .pls explain me sir
Congratulations teacher. You are actually a good instructor. Can I know how you got used of the pencil you are writing with I wish to use it as well in my teaching. I am getting trouble on using it.
Thank you for your inquiry. I use Wacom pen and tablet www.wacom.com/en-us. You get better writing with it when you use it every so often. I think practice make it easier
Thank you so much I never understood this topic and I hated but thank you very much it has become simpler
Thank you so much. We are so happy to hear you are making positive understanding of this topic. Cheers
Hey for the question in 9:07. TO solve DEB, Instead of doing what you did could you also do 180-90-65. Because an angle in a semi-circle = 90 ?
Yes you can. However, the correct order will be 180 - (90 + 25) = 65
Very Helpful vedio Thank you so much Sir
Thanks alot
loved ur video...easy and fun to understand✌🤞
Thanks so much for this
Very informative and conceptual
Glad it was helpful!
was very helpful for revision ,thank u
many thanks for letting me know
Great job sir🎉
thanks alot
Thank you so much .. I learned a lot 🙏🙏🙏
Txs
Wowwwwwww!!!!!! Blessings your way sir !!!!
Thank you so much for finding this video super helpful
Thanks you sir God bless you more 🙏🙏🙏🙏❣️💯
much appreciated
thank you i actually enjoyed the video and it really helped me a lot
👌
You are welcomed
I read some of the comments about this Vedeo and I agree with them that you are an excellent teacher. All the best 👍 and happy new year 🎉
Thank you so much
You make maths easy sir❤
Wow. Thanks so much
44:01 qn 9 how can we subtract 180-134 its not cyclic quadrilateral there is pentagon on circle.
Please respond 😊😢
Thank you Abhishek Oli Chhetri, your observation is absolute right.I had made error there in assuming that was a cyclic quadrilateral.
The video was amazing and helpful I learned so much things😁.
Thanks so much
I have never liked maths teacher but you
Thank you so so much @OsmanNafisah
My novdec maths paper will be on 25th of this month but I just know that I'm going to fail because I had no understanding on maths 🙃
@@OsmanNafisah-eb8qo i pray you do well. Never give up. There is still much that can happen in your favour
Thank you so much 🙏🙏🙏
At 5:54... arent the angles aod and abs both substended by the same arc/are in the same segment so they're both equal?
Thank you so much for the question. it seems so but not quite correct. Angles AOC and ABC are not on the same segment. Infact, angle AOC is not on a segment at all. it just subtends an angle at the centre O.
44:07 why did you say 180-134 when it's not a cyclic quad ? A cyclic quad has 4 points on the circumference but in this case you used a Pentagon , why?
Thank you for spotting this error. You are right.
ua-cam.com/video/GphwjWJ5WwI/v-deo.html
Thank you sir for explaining
thank you so much
45:38 S will be 70 + 23 as it is not cyclic quadrilateral but a pentagon what he is taking
Thanks for spotting that error in the working. Yes, it was a Pentagon and definitely not a quadrilateral
Much appreciation 💛
Your welcome and also nice explanation sir
Helped me a lot because I didn't understand these
A request from you is please correct it by pinning so that people know it
You made my day Prof
Thanks so much
Thank you so much ,this topic has been a real pain for an ibmyp student like me.
gracias
this was extremely helpful. thank you
Glad it was helpful!
Best teacher 👍💕
Thanks a million
Hi I am a bit confused. For s shouldn’t the answer be 93? 70 + 23? As it is not just the angle opposite 110. At 45:21
Thank you so much for your question. Apologies for replying late. Now i think there are two scenarios playing out here. The first is a cyclic quadriateral wherre the opposite angles are supplementary that is 180 degree. The second is an issoccles triangle where the base angles should be the same. In all those two scenarios i expalined how the various angles can be calculated following the rules
@@ICOMATHS thank you, I see now :)
Hello, I have a concern.
For number 9) where you are asked to find r and s I understand how you got the 23 and the 70 but for s wouldn’t you add the 70 and the 23 seen as angle s is from the quadrilateral and then there is a little part in the triangle?
just got to this question myself and had exactly the same issue - i'll let you know if i figure it out but i'm kinda shite at circles so don't get ur hopes up x
Please can you cite the timeline on the video for me to respond quickly
@@ICOMATHS 43:46
Hello at 22:2 I used the angles round a point and you used the cyclic quad but I got 35
at the timeline cited, i dont think i used the cyclic quadrilateral theorem. perpharps you may cite the exact timeline for me to see exactly where you are referring to. Anyway thank you so much for your feedback. They are appreciated
31:06 angle subtented on a semicircle =90 deg. therefore 90+2i=180
2i=180-90 -----> 2i=90 --->i=90/2 =45deg.
sir I'm not sure. is my answer correct at this point....
So sorry i am responding a bit late, My apologies. Now regarding the time cited for referene, i only see a cyclic quadrilateral being explanied there.
You are the best💕
Thank you so much
Thank you soo much you were very helpful ❤️😭✨
Thanks so much for letting us know. We appreciate your feedback
There is therum. Angle subtended by an arc of circle at its centre is double the angle subtended by the Sam arc at remaking circumference. X = 110 Degrees and angle at C = 70 degrees
Yes, you are right
This video is so helpful
Thank you so much
You good Bruh!🤞
thanks so much
Good teacher
😮❤
Thanks
In 24:01 you did 180-36 but it was a minor angle which is less than 180 can you pls explain this to me¿
ok thanks for your question. You see, why i subtracted 36 from 180 and then halfed the result was because the triangle is an isoccesles triangle. The base angles of this type of triangle should be the same.
I hope i explained better for your understanding
Hello, I don’t quite understand why you divided the 304
Degrees by 2, it would be much appreciated if you could reply to this comment and explain, thanks.
Additionally I would like to point out that you could also use the circle theorem that states that the angle subtended(created) at the center by the angle at the circumference in the same segment I.e on the same chord is twice the angle at the circumference. This simply means that you could’ve divided 56 degrees by two and you would arrive at the same 28 degrees.
Thanks for reading my comment and thank you for the vid.
@deanrsingh2398 Thanks so much for the question. Please cite the timeline on the video so I may respond