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A bit faster: Triangles OAD and ODC are isosceles and equal, and triangle OCB is isosceles. Therefore, the angles OAD + ODA + ODC +OCD = 4*67°, and the angles OCB = OBC = X. The sum of the angles of the quadrilateral ABCD = 360°, therefore, 360° = 4*67°+ 2*X → X = 46°.
At a quick glance: 4 sided figure then interior angles sum to 360. OAD and ODC are equal isosceles triangles then ODA , ODC and OCD angles = 67. Angle AOD =180-(2 *67) = 46 = Angle DOC. Then angle COB = 180 - 92 = 88. OCB is isosceles 2*x= 180-88. x= 46 degrees.
Or we can use cyclic quadrilateral theorem to calculate angle DCB. It's 113 degrees . Now let's draw a segment from A to C. We divided the quadrilateral into two triangles. Triangle ABC is a right triangle because one of its sides is at the same time the diameter of the circle. So triangle ADC has two 23 degrees angles and one 134 degrees angle. Now we can use quadrilateral theorem again. It is 134 degrees + x = 180 degrees. Therefore x is equal to 46 degrees.
The four sided figure is made up of three isosceles triangles because the equal sides are radii. This enables a transfer of information to the last triangle which has two equal angles who's sum is 180-88 = 92. 92/2=46 x=46 Solution by diagram. Great!
2:50 I proved that angle ODC had to equal angle AOD because triangles AOD and COD were congruent side-side-side, thus inadvertently proving the angles subtended by the same chord length theorem (which I had forgotten!), and then continued with the angles on a straight line.
A shorter solution: Erect a tangent (T) at A, The angle TAD = 90 - 67 = 23, The angle DAC subtends a similar chord and thus is equal to TAD. The angle ACB = 90, The rest is arithmetic.
ACB is always a right angle. DO is perpendicular to AC so it is parallel to CB. Thus angle AOD is the same as OBC. You only need to calculate AOD. X is AOD.
Another method using central angles and angles subtended by an arc on the circle, using the following theorem: _"The angle subtended by an arc of a circle at its center is twice the angle it subtends on the circle’s circumference."_ First we calculate central angle DOB using theorem above: ∠DOB = 2 × ∠DAB = 2 x 67° = 134° Next we calculate central angle AOD using the fact straight angle AOB = 180° ∠AOD = 180° - ∠DOB = 180° - 134° = 46° Since chords AD and DC are congruent, the so are the central angles they subtend: ∠DOC = ∠AOD = 46° ∠AOC = ∠AOD + ∠DOC = 46° + 46° = 92° Finally, we calculate angle ABC subtended by arc AC using central angle AOC (using theorem above): *∠ABC = x = 1/2 × ∠AOC = 1/2 × 92° = 46°*
By the first method, once you have established that angle AOC is 2x46*, you can conclude because it's the central angle that intercepts the same chord as the inscribed angle ABC. So this latter is half AOC. Sorry for my bad english !
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There is a third method, once you know that angle AOC is 92, you can already say that angle ABC is AOC/2.
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Yet another way to look at it:
Draw line BD. BD bisects x because chords AD and DC are equal. By Thales theorem,
You may use the exterior angle theorem when you add the centre 'o' with the vertex 'c'.
Thus you can conclude
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tri OAD and tri ODC are congruent. radius and equal side so
A bit faster: Triangles OAD and ODC are isosceles and equal, and triangle OCB is isosceles. Therefore, the angles OAD + ODA + ODC +OCD = 4*67°, and the angles OCB = OBC = X. The sum of the angles of the quadrilateral ABCD = 360°, therefore, 360° = 4*67°+ 2*X → X = 46°.
At a quick glance: 4 sided figure then interior angles sum to 360. OAD and ODC are equal isosceles triangles then ODA , ODC and OCD angles = 67. Angle AOD =180-(2 *67) = 46 = Angle DOC. Then angle COB = 180 - 92 = 88. OCB is isosceles 2*x= 180-88. x= 46 degrees.
Or we can use cyclic quadrilateral theorem to calculate angle DCB. It's 113 degrees . Now let's draw a segment from A to C. We divided the quadrilateral into two triangles. Triangle ABC is a right triangle because one of its sides is at the same time the diameter of the circle. So triangle ADC has two 23 degrees angles and one 134 degrees angle. Now we can use quadrilateral theorem again. It is 134 degrees + x = 180 degrees. Therefore x is equal to 46 degrees.
Done the same way
The four sided figure is made up of three isosceles triangles because the equal sides are radii. This enables a transfer of information to the last triangle which has two equal angles who's sum is 180-88 = 92. 92/2=46 x=46 Solution by diagram. Great!
Thanks for video.Good luck sir!!!!!!!!!!!!
So nice of you
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Amazing👍
Thanks for sharing😊😊
Extremely beautiful ❤️🙏 sir as always ❤️🙏🙏🙏🙏
ABCD is a Cyclic Quadrilateral, with opposite corner angles adding to 180°.
X + 2*67 = 180
X = 46°
Excellent!
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Good Morning Master
Thank you Very much for the instruction
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Join BD. In rt triangle ADB, angle ABD = 23°. AD and CD subtends equal angles at B. Hence, x = 2*angleABD = 2*23° = 46°.
thank you sir, for your nice video.
Sir if you make a video with "indeterminate form" it will benefit many people.
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Welcome sir.
2:50 I proved that angle ODC had to equal angle AOD because triangles AOD and COD were congruent side-side-side, thus inadvertently proving the angles subtended by the same chord length theorem (which I had forgotten!), and then continued with the angles on a straight line.
I have solved it by second methed. But now I know the 1st method too thanks my HERO
Thanks it really helped😊❤
Join AC & DB.
Angle ADB = 90
Angle DAB = 67
Hence Angle ABD = 23
Angle ACD = 23 angles subtended by chord AD.
In triangle ADC the DAC = ACD= 23
Hence angle ADC =134
Since ABCD is cyclic quadrilateral
ADC + x = 180
134+x = 180
X = 46
A shorter solution: Erect a tangent (T) at A, The angle TAD = 90 - 67 = 23, The angle DAC subtends a similar chord and thus is equal to TAD. The angle ACB = 90, The rest is arithmetic.
ACB is always a right angle.
DO is perpendicular to AC so it is parallel to CB.
Thus angle AOD is the same as OBC.
You only need to calculate AOD. X is AOD.
Sir can u please xplain how u did COD=46°
2:50 because the triangles AOD and COD are congruents.
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I am obliged
Maths rocks
Another method using central angles and angles subtended by an arc on the circle, using the following theorem:
_"The angle subtended by an arc of a circle at its center is twice the angle it subtends on the circle’s circumference."_
First we calculate central angle DOB using theorem above:
∠DOB = 2 × ∠DAB = 2 x 67° = 134°
Next we calculate central angle AOD using the fact straight angle AOB = 180°
∠AOD = 180° - ∠DOB = 180° - 134° = 46°
Since chords AD and DC are congruent, the so are the central angles they subtend:
∠DOC = ∠AOD = 46°
∠AOC = ∠AOD + ∠DOC = 46° + 46° = 92°
Finally, we calculate angle ABC subtended by arc AC using central angle AOC (using theorem above):
*∠ABC = x = 1/2 × ∠AOC = 1/2 × 92° = 46°*
Angle ADC= 67•2 = 134
x= 180-134=46
AD=DC and OA=OD=OC=Radius → ∆AOD=∆DOC are isosceles → ∠DAO=67º =∠ADO=∠CDO=∠DCO → ∠AOD=∠DOC=180º - 67º -67º=46º →→→ ∠ABC =X subtends AC → Corresponding central angle= 2X =∠AOC = ∠AOD+∠DOC= 46º+46º → X=46º
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DAB = 67° so DCB = 180° - 67° = 113°.
AB is a diameter so ACB = 90°
so ACD = 113° - 90° = 23°.
Also CAD = ACD = 23°.
ADC = 180° - 2x(23°) = 134°.
ABC = 180° - ADC = *46°*
That was easy
thnku
OD parallel BC
X=46° corresponding angles
ADB=90
So ABD=90-67=23
Then ABD=DBC=23
Therefore, ABC=46
x is 46 degree by using quadrilateral angle sum and cylic theorm
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By the first method, once you have established that angle AOC is 2x46*, you can conclude because it's the central angle that intercepts the same chord as the inscribed angle ABC. So this latter is half AOC. Sorry for my bad english !
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Opposite exterior angles is a third solution.
x = (360 - 4 * 67°) / 2 = 46°
👍.
Thales rings a bell ...
180-67*2 = 46
46
x=46
X=46deg.
X=23
X=46. Am I right?
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