CSEC Maths - Circle Geometry Past Papers

In this video I go through some past papers on Circle Geometry with my online CSEC Maths Class.
Online CSEC Maths, Chemistry, Physics Lessons
868-784-2059

[@leahbishop1117]

I watched a lot of circle theorem videos & this video is the first one that I can say that I completely understand circle theorem.

[@CSECMathematics]

Excellent.

[@MontellJacobs-o4w]

for reallllll

[@MutintaChitendele]

Good explaination sir

[@jaden2446]

Thank you so much sir for helping us. By far one of the best teachers out there.

[@relaxgirl5129]

Totally agreed

[@kellyyy1333]

Facts

[@clipville00]

Anyone here in 2024? 😅

[@5thdimentionalbeing367]

Yessir

[@SusanDomingo-f7e]

😂here...private candidate

[@marianewman8638]

Ma ppl (ಥ﹏ಥ)

[@Dxspair.]

ofc😂

[@neikahannam8385]

Yey

[@LEOYT_BABBY]

Exam in 2 days paper 2😢 anyone else?

[@abayomisylvester6838]

same

[@VasileFlorinSenciuc]

min. 1:07:05
BOE=360-BOE: =360-140=220(REFLEX ANGLE), Then BFE is 1/2 (220) =110 A cord makes 2 segment in a circle. Remember you can apply the theorem of 'angle at the centre is twice the angle at the circumference,' for minor and major arcs as well.

[@silas533]

Can someone explain what will be the reason for OAF for the 2013 question

[@TheNikbel]

For question 1(iii) I used the Alternate Segment Theorem to get my answer which was the same 32. Would it be corrects in terms of reasoning?

[@jaden2446]

yes it would be correct

[@lenicebethelmy7648]

I was thinking that same thing sir Thank You!

[@jeleelad.4021]

Thanks sir, this was extremely helpful.

[@CSECMathematics]

Glad it helped

[@JellyJay.]

thank you so much I feel much more confident for these type questions🤭👏

[@isatujalloh552]

Thank you very much it really helped me a lot

[@CSECMathematics]

Glad it helped

[@bensonnyondo8691]

helpful... thanks

[@Caribbean_Doll500]

how do i access the papers thank you

[@sitisunny6058]

couldn't the reason for JGH be the same as JHE ?(the angle in a semi circle is 90 degrees)

[@ayannalewis8470]

Opposite angles in cyclic quadrilateral are supplementary CMB =180-58=122

[@desiwidthalocs]

thank you so much for this sir, really really helpful

[@PreciousGeorge-uo9nm]

Sir are you doing any revisions for January exams

[@jouvanayholloway4960]

Is there any way that I could get the questions?

[@gaminghellfire]

Ok this is basically O level question which is quite easy

[@ayannalewis8470]

FGH=68 the reason the angle made by a tangent and chord is segment

[@ayannalewis8470]

JHE is the angle between a tangent and. Chord is 90°

[@ayannalewis8470]

Reason the angle in semi circle is 90° , Angle ABC = 180-(58+90)=32°

[@ayannalewis8470]

Angle at the center is twice the angle at circumference

[@tantanwilson6183]

Hello

[@ayannalewis8470]

Opposite angles in a cyclic quadrilateral are supplementary

[@ayannalewis8470]

50 reason the angle of centre is twice on the circumference

[@ayannalewis8470]

The angle between a tangent and chord is same

[@ayannalewis8470]

A chord involve and tangent are the same touching a point A

[@thatguyfrom-yx8pk]

I used alternate segment theorem for may 2013 OAF and got 55.

[@ayannalewis8470]

The angle in a semi circle is 90°

[@ayannalewis8470]

Yes sir the question before you just flip it to get our answer

[@novisheensalinel7264]

Thank u so much 🥰

[@ayannalewis8470]

Opposite angles in a cyclic quadilateral supplementry

[@ayannalewis8470]

The angle in a semi circle Angel AbC = 180- 90+58 = 32

[@myaben4366]

90+58 = 148 180-148 =32

[@bintubarre2680]

I wanna join your class sir

[@ayannalewis8470]

CMB = 180-58=122° my reason the angles are cyclic quadrilateral

[@ayannalewis8470]

Cyclic quadilateral a+c =180 b+d= 180

[@ayannalewis8470]

SOQ=2×58=116° , SO=OQ=radius

[@ayannalewis8470]

There no working NCM is also 32° is right angle triangles

[@ayannalewis8470]

UVY= Angles in the same segment are equal

[@ayannalewis8470]

180-(90+58)=32 and also a right angle triangle

[@ayannalewis8470]

Angel is ABC = 180-(90+58)=32°

[@dinsmorekeir4807]

180-116= 64/2 =32

[@adgf1x]

[@empsiemanual6117]

thank you very much indeed

[@CSECMathematics]

You are very welcome

[@kellybrownnelson2947]

Thank you idk what I would have done 😭

[@adgf1x]

[@ayannalewis8470]

Good morning sir

[@silxbeats]

This is becoming hella easy to do...with practice...

[@ayannalewis8470]

CMN=180-(58+90)=32°

[@ayannalewis8470]

FAB =180-(75+50)=55°

[@ayannalewis8470]

Right angle triangle 90°

[@ayannalewis8470]

NCM is 180(58+90)=32

[@adgf1x]

[@ayannalewis8470]

Afternoon sir

[@bensonnyondo8691]

helpful... thanks

[@ayannalewis8470]

NCM=180-(58+90)=32

[@ayannalewis8470]

OQS= 180-116/2=32 because issoless triangle

[@jeanettelee5608]

thank you sir

[@ayannalewis8470]

Afternoon sir and classmates

[@ayannalewis8470]

CMB is 180-58 = 122

[@ayannalewis8470]

EO =OB radius =180-(20+20)=140

[@andreinadelacruz1022]

[@ayannalewis8470]

RSPQ= cyclic qraudrilateral

[@ayannalewis8470]

SPQ= 180-58=122°

[@r.s9643]

Thanks a lot sir

[@CSECMathematics]

Most welcome

[@francois8422]

Why not drawn to scale? , not good

[@ayannalewis8470]

TSR=180-56=124°

[@1973yogesh]

🙏🙏👍👍

[@ayannalewis8470]

Right angle triangle

[@ayannalewis8470]

SPQ180-58=122

[@ayannalewis8470]

SOQ=2×58= 116 180-116÷2=32

[@ayannalewis8470]

AB is a chord

[@ayannalewis8470]

E,H, F= issoles triangle

[@ayannalewis8470]

Angel 180-58=122

[@ayannalewis8470]

BOa= 180(40+40)100°

[@ayannalewis8470]

ABC 58-90= 32

[@ayannalewis8470]

CMB 180-58=122°

[@ayannalewis8470]

AB= diameter

[@ayannalewis8470]

NCM 180 (58+90)=32

[@ayannalewis8470]

AB=Diameter

[@ayannalewis8470]

180-(90+58)=32

[@ayannalewis8470]

Hearing you sir

[@ayannalewis8470]

180 -(46+46)=

[@ayannalewis8470]

CMB =180-58 =122

[@ayannalewis8470]

180-(46-56)=

[@tomorrow-il7mm]

3 hours before the exam

[@ayannalewis8470]

And=32°

[@ayannalewis8470]

180-44/2=68°

[@tajayfrancis9485]

37:42

[@ayannalewis8470]

Morning sir plus classmates

[@ayannalewis8470]

TPQ=180-(46+32)=102-46=56°

[@ayannalewis8470]

180- 122=58

[@ayannalewis8470]

58- 90= 32

[@reazhoosein7c487]

90degrees sir

[@ayannalewis8470]

180-44÷2=68

[@ayannalewis8470]

BFE=90+20=110

[@ayannalewis8470]

55+35=90°

[@ayannalewis8470]

2×64=128

[@ayannalewis8470]

46°

[@ayannalewis8470]

Diameter

[@ayannalewis8470]

78°

[@ayannalewis8470]

90°