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.
I watched a lot of circle theorem videos & this video is the first one that I can say that I completely understand circle theorem.
Excellent.
for reallllll
Good explaination sir
Thank you so much sir for helping us. By far one of the best teachers out there.
Totally agreed
Facts
Exam in 2 days paper 2😢 anyone else?
same
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.
Anyone here in 2024? 😅
Yessir
😂here...private candidate
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ofc😂
Yey
Opposite angles in cyclic quadrilateral are supplementary CMB =180-58=122
Thanks sir, this was extremely helpful.
Glad it helped
Thank you very much it really helped me a lot
Glad it helped
helpful... thanks
thank you so much I feel much more confident for these type questions🤭👏
Ok this is basically O level question which is quite easy
Reason the angle in semi circle is 90° , Angle ABC = 180-(58+90)=32°
thank you so much for this sir, really really helpful
JHE is the angle between a tangent and. Chord is 90°
Angle at the center is twice the angle at circumference
Hello
I used alternate segment theorem for may 2013 OAF and got 55.
Yes sir the question before you just flip it to get our answer
FGH=68 the reason the angle made by a tangent and chord is segment
SOQ=2×58=116° , SO=OQ=radius
The angle in a semi circle Angel AbC = 180- 90+58 = 32
90+58 = 148 180-148 =32
Opposite angles in a cyclic quadrilateral are supplementary
Opposite angles in a cyclic quadilateral supplementry
Can someone explain what will be the reason for OAF for the 2013 question
180-116= 64/2 =32
Thank u so much 🥰
Cyclic quadilateral a+c =180 b+d= 180
180-(90+58)=32 and also a right angle triangle
thank you sir
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?
yes it would be correct
I was thinking that same thing sir Thank You!
Angel is ABC = 180-(90+58)=32°
thank you very much indeed
You are very welcome
OQS= 180-116/2=32 because issoless triangle
50 reason the angle of centre is twice on the circumference
The angle in a semi circle is 90°
Sir are you doing any revisions for January exams
Afternoon sir
Afternoon sir and classmates
EO =OB radius =180-(20+20)=140
CMB = 180-58=122° my reason the angles are cyclic quadrilateral
UVY= Angles in the same segment are equal
A chord involve and tangent are the same touching a point A
Thank you idk what I would have done 😭
Right angle triangle 90°
helpful... thanks
NCM is 180(58+90)=32
Good morning sir
CMN=180-(58+90)=32°
FAB =180-(75+50)=55°
RSPQ= cyclic qraudrilateral
NCM=180-(58+90)=32
Right angle triangle
🙏🙏👍👍
This is becoming hella easy to do...with practice...
couldn't the reason for JGH be the same as JHE ?(the angle in a semi circle is 90 degrees)
Hearing you sir
Thanks a lot sir
Most welcome
E,H, F= issoles triangle
SPQ= 180-58=122°
CMB is 180-58 = 122
180-(90+58)=32
how do i access the papers thank you
TSR=180-56=124°
Is there any way that I could get the questions?
NCM 180 (58+90)=32
I wanna join your class sir
SOQ=2×58= 116 180-116÷2=32
The angle between a tangent and chord is same
SPQ180-58=122
AB is a chord
AB=Diameter
BOa= 180(40+40)100°
AB= diameter
Angel 180-58=122
180 -(46+46)=
CMB 180-58=122°
ABC 58-90= 32
180-(46-56)=
CMB =180-58 =122
TPQ=180-(46+32)=102-46=56°
There no working NCM is also 32° is right angle triangles
And=32°
180-44/2=68°
55+35=90°
2×64=128
3 hours before the exam
46°
58- 90= 32
180- 122=58
78°
Morning sir plus classmates
BFE=90+20=110
180-44÷2=68
Why not drawn to scale? , not good
37:42
90°
90
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