A Nice Geometry Problem | You should be able to solve this! | 3 Methods
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- čas přidán 8. 07. 2024
- A Nice Geometry Problem | You should be able to solve this! | 3 Methods
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Method #1:
There are two rules which govern this method:
1) The angle of an arc when measured at the circumference is half the angle of the same arc when measured at the center.
2) The internal angles of a polygon always sum to (n-2)180°, where n is the number of sides of the polygon.
By 1), ∠CAB and ∠COB cover the same arc CB. As ∠COB = θ and O is the center of the circle, ∠CAB = θ/2, as A is on the circumference.
By 2), as ABOC is a quadrilateral with n=4 sides, the internal angles of ABOC sum to (4-2)180° = 360°. This means that ∠ABO, ∠BOC (note that this is the _internal_ angle, so it's not θ but rather 360°- θ), ∠OCA, and ∠CAB sum to 360°.
20° + (360°- θ) + 20° + θ/2 = 360°
40° + 360° - θ + θ/2 = 360°
θ/2 = 40°
θ = 2(40°) = 80°
Method #2:
Draw OA, then draw OD, so that AD is a diameter. As OA, OB, OC, and OD are all radii of circle O, they are all equal. From this we can see that ∆BOA and ∆AOC are both isosceles triangles, as OA = OB and OA = OC. That means that if ∠ABO = α, then ∠OAB = α as well, and if ∠OCA = β, then ∠CAO = β as well.
This means that, as ∠DOB is an exterior angle to ∆BOA at O, ∠DOB = α+α = 2α, and as ∠COD is an exterior angle to ∆AOC at O, ∠COD = β+β = 2β. As ∠DOB+∠COD = ∠COB = θ, then θ = 2α+2β = 2(α+β). α and β each equal 20°, so:
θ = 2(α+β) = 2(20°+20°) = 2(40°) = 80°
20 + 20 + (360 - θ ) + θ/2 = 360 ( Arrowhead interior angles) ====> θ = 80
θ=α+α→ Ángulos: AOB=180-α→ BAO=180-20-180+α=α-20→ Ángulo central =2*BAO=2α-40=α→ α=40º→ θ=2*40=80º.
Gracias y saludos.
I used the 1st method but from 2:00, I just got angle BAC = 20°+20°= 40°, hence angle theta (I.e BOC) is 2 * angle BAC = 2 * 40° = 80°
In the Method 1
Extend AO to D
Triangle AOB
Exterior angle BOD =20+20=40 degrees --(1)
Triangle AOC
Exterior angle COD
= 20+20=40 degrees -- (2)
By (1) + (2) we get
Angle BOD + angle COD =80 degrees
BOC =80 degrees
angle BAC=1/2 angle BOC as they both have the same chord and one is a central angle ( BOC) , another is inscribed.
The second methods and third methods were identical I have noticed due the same central point being used. And as for me the first method seemed more intuitive.
BO -> B'; CO -> C';
80
Let's label the angle at the circumference n (below A), then theta = 2n as the angle at the circle's center is twice the angle at the circumference (circle theorem).
2n + O = 360 equation 1 (since both add ups to a complete revolution or 360 degrees)
O = 360- 2n equation 2
n + O + 20 + 20 = 360 (since the internal angle of a quadrilateral =360)
n+ O = 360-40
n+ O = 320 equation 3
n + 360- 2n=320 substitute equation 2 into equation 3
-n = -40
n= 40
Hence , 2n =80 = theta Answer
Please remember 2n= theta. See above
OA=OB=R --> OAB=ABO=20. OA=OC=20 --> OAC=OCA=20. Now BAC=OAB+OAC=20+20=40. Thus BOC = 2*BAC = 2*40 = 80
{20°A+20°B+20°C}= 80°ABC{ 360°/80°ABC}=40.40 2^20.2^20 1^5^4.1^5^4 1^12^2.1^12^2 1^1.1^2 1^2 (ABC ➖ 2ABC+1)
I solved this orally😂❤
Geometria é como um lazer, satisfação total!
θ/2+20+20+(360-θ)=360...θ=80
Theta=2A°
40+360-2A°+A°=360
A°=40
Theta=2(40)=80°
20+20+ ½(theta)+(360-theta)=360°
φ = 30° → θ = 8φ/3
Very simple
360-140-140=80
Why you prefer tedious method ..
Delta==80°
@ = 80
🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉
80 degrees
❤❤❤❤❤❤❤❤❤❤❤❤
Join AO & BC
Triangle BOC is isosceles as BO = CO
Then ang OBC = ang OCB
As angle BAC =40
then ang ABC + ang ACB
= 180-40=140
ang OBC + ang OCB
=140 - 20-20=100
Then
BOC = 180 - 100=80
Please see another method that I have already offered
это устная задача, придуравчок