I solved it using the property of the exterior angle of a circumference (not of a triangle), for which I also had to draw the segment OB: 22º=(x-22º)/2; then x=66º. Thanks!
Awesome! I'm thinking the two premises needed to form a conclusion here are: Exterior Angle Thm and Isosceles angle relationship. Thank you for reinforcing critical thinking. 🙂
X = 66 degrees. ABO is an isosceles triangle (OC is the Radiusses is thus also OB). The angle OBC is 44 (exterior angle theorem), which renders the angle BOC as 180-44-44=92. Thus x is 180-92-22=66.
triangle ABO: angles in pts. A and O = 22°, angle in pt. B = 136°. Then in triangle BCO: angles in pts. B and C = 44° and angle in O = 92°. Thus x = 180° - 92° - 22° = 66°
I didn't know the "Exterior Angle Theorem". However, it makes sense, of course, and the solving of the problem a bit shorter since you don't have to find the missing angles of all the triangles. :)
I solved it drawing a specular triangle to A0C. Some minutes lost, anyway I 'm happy of my progress in math. Thanks for your lessons: 3 months ago I didn' t know what an exterior angle was!
another approach using intersecting secant properties. Let AO cut the circle at D and E forming two sectors on circle arc BD...smaller and arcCE ... Larger the property is angle OAC = 1/2 (Larger arc- Smaller arc) ; 1/2(arcCE-arcBD) arc CE = x and arc BD = 22 ( you arrived at) also OAC = 22 given therfore angle OAC 22 = 1/2(x-22) x - 22 = 44 x = 66
LOL The thumbnail omitted AB=OC! I tried to solve it before starting the video and wasted 10 minutes breaking my brain over how the damn thing could ever be solved. xD
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I solved it using the property of the exterior angle of a circumference (not of a triangle), for which I also had to draw the segment OB: 22º=(x-22º)/2; then x=66º. Thanks!
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you are great sir.......what a way of explanation to such an
incredible problem
Wow
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Awesome! I'm thinking the two premises needed to form a conclusion here are: Exterior Angle Thm and Isosceles angle relationship. Thank you for reinforcing critical thinking. 🙂
Good point!
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@@PreMath I guessed that the initial diagram having (2 identical sides) would give the next 2 angles. I like the way U did it better.
X = 66 degrees. ABO is an isosceles triangle (OC is the Radiusses is thus also OB). The angle OBC is 44 (exterior angle theorem), which renders the angle BOC as 180-44-44=92. Thus x is 180-92-22=66.
I did the same
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triangle ABO: angles in pts. A and O = 22°, angle in pt. B = 136°. Then in triangle BCO: angles in pts. B and C = 44° and angle in O = 92°. Thus x = 180° - 92° - 22° = 66°
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I didn't know the "Exterior Angle Theorem". However, it makes sense, of course, and the solving of the problem a bit shorter since you don't have to find the missing angles of all the triangles. :)
Amazing way of teaching👍
Thank you so much for your hard work😊😊
angle x = angle OAC + angle ACO
= angle OAC + angle CBO
( since ∆ BOC is isosceles, OB= OC)
= 2 angle OAC + angle ABO
= 3 angle OAC = 66°
( since ∆ ABO is isosceles,
as AB= OC = OB)
😁"Isosceles triangle " is the key point!! Fantastic solution!!
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Solved it in a similar way, just by using sum of angles in triangle and straight line.
I solved it drawing a specular triangle to A0C. Some minutes lost, anyway I 'm happy of my progress in math. Thanks for your lessons: 3 months ago I didn' t know what an exterior angle was!
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V nice circle question with exterior angle playing an important role
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\
amazing video, I watch this video with full concentration
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another approach using intersecting secant properties.
Let AO cut the circle at D and E forming two sectors on circle arc BD...smaller and arcCE ... Larger
the property is angle OAC = 1/2 (Larger arc- Smaller arc) ; 1/2(arcCE-arcBD)
arc CE = x and arc BD = 22 ( you arrived at) also OAC = 22 given
therfore angle OAC 22 = 1/2(x-22)
x - 22 = 44
x = 66
👍👍👍👍more geometry problem please and thanks!
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You are always the best thinks a lot for your helps
How do you know that ABO is isosceles triangle??
sure can, thanks for the challenge bro
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I like the problem explanation very much
Extremely beautiful sir
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Nice. Very good indeed.
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Join OB. Now AB=OC=OB.
So Angle OBC=AngleOCB=2×22= 44 degrees.
So the required angle x=44+22= 66 degrees.
Very nice geometry question!!! My exams are there so I will not be able to check up your videos
Best of luck on your exams
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I got 66 degrees but by a different route:
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Hmm, I didn't catch that AB = OC.
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I had no idea abt exterior opposite angle theorem
I.have been watching your videos for six months
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Yes...i am able to do this...😊
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@@PreMath Thank you sir😊❤️
LOL The thumbnail omitted AB=OC! I tried to solve it before starting the video and wasted 10 minutes breaking my brain over how the damn thing could ever be solved. xD
Sum of 345 is 12
But here is 216 that is 12×18
Multiplier is 18
3×18=54
4×18 =72
5×18=90
Mukundsir
Couldn't solve it just by looking at the thumbnail because the thumbnail doesn't show those two line segments being of equal length. lol
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66 degrees. Good question it was.
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nice
Super
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Solved easily
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It is easy to prove that x is the triple of the first angle !
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no more 1 min, X = 66
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wow!!
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Great Great Great
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Suport from me
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70.... Anzi 66
1/2x=22
X=44
Wow
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Approve
Easy
66
Great
Super duper easy ans: x = 66 degree
The answer is x = 66 degree lol
Cross check my friend
@@Teamstudy4595 yes ans is x=66 degree but if line ac would have touched the outer side of circle then it would've x=44 degree