Can you calculate the angles a, b, and c? | (Circle) |

[@SladeMacGregor]

This problem is awesome PreMath!

[@PreMath]

Glad to hear that! Thanks for the feedback ❤️

[@davidellis1929]

Another way to solve it is by finding the arc ABC to be twice angle D, or 228 degrees. Since arc AB is 142 degrees, that leaves arc BC at 86 degrees, so angle A is half 86+100, or 93 degrees. From here, one can say angle C is 180-93, or 87 degrees, and angle B is 180-114, or 66 degrees.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@semeemaazad8948]

Yes. I found this way. Sir showed another way thank u👍🏼👍🏼

[@alster724]

Thanks for the refresher on arc measures.

[@PreMath]

Glad it was helpful! 👍 Thanks for the feedback ❤️

[@PrithwirajSen-nj6qq]

Reflex angle AOC=114*2=228 degs Arc BC =228 -142=86 degs Arc AD = 360-228-100=32 degs c=(142+32)/2=174/2 =87 degrees b=(100+32)/2=66 degrees a=(100+86)/2=93 degrees

[@PreMath]

Excellent! Thanks for sharing ❤️

[@quigonkenny]

As quadrilateral ABCD is a cyclic quadrilateral, opposite angles must sum to 180°. As angle d is given as 114°, then [ b = 180°-114° = 66° ] As the angle subtending an arc from the circumference is half the angle subtending the same arc from the center, then as ∠ADC (angle d) = 114°, ∠AOC = 2(114°) = 228°. As ∠AOB = 142°, ∠BOC = 228°-142° = 86°. As ∠CBA (angle b) = 66°, then ∠COA = 2(66°) = 132°. This also could have been determined by subtracting the value of ∠AOC (228°) from 360°. As ∠COD = 100°, then ∠DOA = 132°-100° = 32°. As angle a subtends arcs CD and BC, then: a = (∠COD+∠BOC)/2 a = (100°+86°)/2 [ a = 186°/2 = 93° ] As angle c subtends arcs AB and DA, then: c = (∠AOB+∠DOA)/2 c = (142°+32°)/2 [ c = 174°/2 = 87° ]

[@PreMath]

Excellent! Thanks for sharing ❤️

[@ChuzzleFriends]

ABCD is a cyclic quadrilateral, so opposite angles must be supplementary. Therefore, m∠B = b° = 180° - 114° = 66°. (b = 66) Let's draw radii AO & CO. This forms a central ∠AOC which subtends the same arc with ∠B. By the Measure of an Inscribed Angle Theorem, m∠AOC = 66° * 2 = 132°. So, minor arc AC must also measure 132° (A measure of a central angle is the measure of the arc it subtends). Since the minor arc CD already measures 100°, arc AD must be a minor arc that measures 32°. Draw radii BO & DO. This forms another central ∠BOD which subtends the same arc with ∠C. By the Arc Addition Postulate, arc BD must be a minor arc that measures 142° + 32° = 174°. So, m∠BOD = 174°. By the Measure of an Inscribed Angle Theorem, m∠C = c° = (174°)/2 = 87°. (c = 87) Finally, m∠A = a° = 180° - 87° = 93°. (a = 93) So, the values of the variables are as follows: a = 93 b = 66 c = 87

[@PreMath]

Excellent! Thanks for sharing ❤️

[@matthewmcdaid7962]

I solved it using isosceles triangles. Four triangles with apexes at O. And of course got the same result.

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@RealQinnMalloryu4]

{142°AB+114DC+100°O}=356°ABDCO/ 360°/356°ABDCO=1.4ABDCO 4:21 1.2^2 1.1^2 1.2 (ABDCO ➖ 2ABDCO+1).

[@PreMath]

Thanks for the feedback ❤️

[@neilmorrone691]

Very nice explanation of this lesson on a topic in the subject of Geometry. Grazie!

[@PreMath]

Glad to hear that! You are very welcome! Thanks for the feedback ❤️

[@jamestalbott4499]

Thank you!

[@PreMath]

Excellent! You are very welcome! Thanks for the feedback ❤️

[@sorourhashemi3249]

Thanks. Easy

[@PreMath]

I'm glad you found it easy! 👍 Thanks for the feedback ❤️

[@misterenter-iz7rz]

b=180-114=66, 114-(142/2)=43, angle BC=43×2=86, thus angle AD=360-100-142-86=32, a=86/2+100/2=93, thus c=180-93=87.😊

[@PreMath]

Excellent! Thanks for sharing ❤️

[@marioalb9726]

b = 180° - 114° = 66° Angle AOC = 2*66°= 132° Angle AOD = 132°-100°= 32° c = ½ (142°+32°) = 87° a = 180° - c = 93° ( Solved √ )

[@PreMath]

Excellent! Thanks for sharing ❤️

[@highlyeducatedtrucker]

Another way to do this one: Draw an auxilliary line from A to C. Then you can use the Central Angle Theorem to find angle ACB and angle CAD. (Angle ACB is 71, angle CAD is 50.) Then use sum of angles in a triangle = 180 to find out the rest. (Angle ACD = 180-114-50=16. Angle BCD = Angle ACD + Angle ACB = 71 + 16 = 87. And Angle CAB = 180 - Angle BCD = 180 - 87 = 93.)

[@PreMath]

Excellent! Thanks for sharing ❤️

[@Birol731]

My way of solution ▶ ∠ADC = 114° Since the sum of opposite angles of a quadrilateral inscribed in a circle is 180°, we can write: ∠ADC + ∠CBA = 180° ⇒ ∠CBA = 66° b= 66° ✅ ∠ADC= 114° 114°*2= ∠(AB) + ∠(CB) ∠(AB)= 142° ⇒ 228°= 142° + ∠(CB) ∠(CB)= 86° a= [∠(DC) + ∠(CB)]/2 ∠(DC)= 100° ∠(CB)= 86° ⇒ a= 93° ✅ a+c= 180° a= 93° ⇒ c= 87° ✅ a= 93° b= 66° c= 87°

[@PreMath]

Excellent! Thanks for sharing ❤️

[@johankotze42]

I initially thought PreMath had lost his marbles when he first explained the subtended angle using AOB etc. 😀 I did the solution by drawing in a;; the radii from O to A B C and D. Then you end up with a bunch of isosceles triangles.

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@unknownidentity2846]

Let's find the angles: . .. ... .... ..... A convex quadrilateral ABCD is cyclic if and only if its opposite angles add up to 180°. So we can conclude: d = ∠ADC = 114° ⇒ b = ∠ABC = 180° − d = 180° − 114° = 66° According to the central angle theorem we obtain: ∠ACB = ∠ADB = ∠AOB/2 = 142°/2 = 71° ∠CAD = ∠CBD = ∠COD/2 = 100°/2 = 50° ∠BDC = ∠ADC − ∠ADB = 114° − 71° = 43° Let E be the point where AC and BD intersect. Now we apply the exterior angle theorem to the triangle ADE: ∠CED = ∠DAE + ∠ADE = ∠CAD + ∠ADB = 50° + 71° = 121° From the interior angle sum of the triangle CDE we obtain: 180° = ∠DCE + ∠CDE + ∠CED = ∠ACD + ∠BDC + ∠CED = ∠ACD + 43° + 121° = ∠ACD + 164° ⇒ ∠ACD = 180° − 164° = 16° Now we are able to calculate the values of the remaining angles: c = ∠BCD = ∠ACD + ∠ACB = 16° + 71° = 87° a = ∠BAD = 180° − c = 180° − 87° = 93° Best regards from Germany

[@PreMath]

Excellent!😀 Thanks for sharing ❤️

[@LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL : 01) Arc AB = 142º 02) Arc ABC = (114º * 2) = 228º 03) Arc BC = 228º - 142º = 86º 04) Arc CD = 100º 05) Arc AD = 360º - (142º + 86º + 100º) = 360º - 328º = 32º 06) Angle a = (186º / 2) = 93º 07) Angle b = (132º / 2) = 66º 08) Angle c = (174º / 2) = 87º 09) Angle d = 114º 09) Check : 93º + 66º + 87º + 114º = 360º Thus, OUR BEST ANSWER IS : Angle a = 93º Angle b = 66º Angle c = 87º

[@PreMath]

Excellent! Thanks for sharing ❤️

[@gelbkehlchen]

Solution: In a chord quadrilateral, the opposite angles add up to 180°. Therefore: b+114° = 180° |-114° ⟹ b = 66° The corresponding central angle at 114° is twice as large, namely 228°. Therefore: arc BC + 142° = 228°|-142° ⟹ arc BC = 86° ⟹ Central angle at a = arc BC + arc CD = 86°+100° = 186° ⟹ a = 186°/2 = 93° ⟹ c = 180°-93° = 87°

[@PreMath]

Excellent! Thanks for sharing ❤️

[@PrithwirajSen-nj6qq]

Ang DOC =100 degs Ang ODC =ang OCD =(180-100)/2=40 degs Ang AOC=142 degs Ang OAC =ang OCA =(180-142)/2=19 degs Any b =114/2=66 degs Ang OBC =angle OCB =66-19=47 degs c=40+47=87 degs Angle a=360-66-87-114=93 degs

[@PreMath]

Excellent! Thanks for sharing ❤️

[@marcgriselhubert3915]

Fine.

[@PreMath]

Thanks for the feedback ❤️

[@DaRealNoobKing]

stop not using vectors, pls just use vectors

[@AmirgabYT2185]

93°; 66°; 87°

[@PreMath]

Excellent! Thanks for sharing ❤️

[@wackojacko3962]

@ 4:13 , how does that Central Angle Theorem work when Central Angle is 0⁰ ? It's a bizarre lookin Bowtie too say the least. 🤔 A Bowtie with only two lines! 😊

[@PreMath]

Wow😀 Thanks for sharing ❤️

[@PrithwirajSen-nj6qq]

Join AC Ang CAD = 1/2 ang COD =50 Now from 🔺 ACD ang ACD =180-114 - 50=16 Angel ACB =1/2*ang AOC = 142/2=71 Hence c =16+71 =87 a= 180-87 =93 b= 180- 114=66

[@PreMath]

Excellent! Thanks for sharing ❤️

[@TramHicks]

I understand some

[@PreMath]

No worries... Thanks for the feedback ❤️

[@k9slayer]

magic

[@PreMath]

Excellent! Glad to hear that! Thanks for the feedback ❤️