A Nice Geometry Problem | You should be able to solve this! | 3 Methods
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- เผยแพร่เมื่อ 5 ก.ย. 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
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
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°
BO -> B'; CO -> C';
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 is 40 deg because it is made of 2 20deg angles of isos triangles AOB and AOC. So theta = 80 deg, angle at centre = 2x angle at circum.
angle BAC=1/2 angle BOC as they both have the same chord and one is a central angle ( BOC) , another is inscribed.
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.
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
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.
Theta=2A°
40+360-2A°+A°=360
A°=40
Theta=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)
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
Very simple
360-140-140=80
Why you prefer tedious method ..
Geometria é como um lazer, satisfação total!
θ/2+20+20+(360-θ)=360...θ=80
20+20+ ½(theta)+(360-theta)=360°
φ = 30° → θ = 8φ/3
Delta==80°
@ = 80
I solved this orally😂❤
80 degrees
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Far too complicated. Just draw a line from the centre of the circle to the top point. We've now got two bilateral triangles, the bottom angles of which are 20 degrees, so the top angle is twice that, namely 40 degrees. (He did this, but it took over two minutes.) Then since the angle subtended by a chord at the centre is twice that subtended by the same chord at any point on the circumference, 𝚹 is twice 40, ie. 80 degrees. Some of his problems are interesting, but the presentation is painfully slow.
это устная задача, придуравчок