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.
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° ]
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
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.)
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
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.
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°
@ 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! 😊
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
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
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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.
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Thanks for the refresher on arc measures.
I solved it using isosceles triangles. Four triangles with apexes at O. And of course got the same result.
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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
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Very nice explanation of this lesson on a topic in the subject of Geometry. Grazie!
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stop not using vectors, pls just use vectors
b = 180° - 114° = 66°
Angle AOC = 2*66°= 132°
Angle AOD = 132°-100°= 32°
c = ½ (142°+32°) = 87°
a = 180° - c = 93°
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{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).
93°; 66°; 87°
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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° ]
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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
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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.)
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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.😊
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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
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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.
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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°
@ 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! 😊
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Let's find the angles:
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..
...
....
.....
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
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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
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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
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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°
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magic
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