CIRCLE 2

(Angles at Circle)

MGMP Matematika

SMPK PENABUR Jakarta

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After learning this topic, students are able to :

Learning Achievement

• Determine the relationship between angle at center (central angle and angle on circumference (inscribed angle) which subtended the same arc.

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• Solving real problem related to central angle and inscribed angle.

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Material of circle 2 :

• Angle properties of a circle
• Angles on cyclic quadrilateral
• Angles of intersecting chords
• Angles of intersecting secants
• Ptolemy Theorem
• Angle n-sided polygon

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PowerPoint Presentation

Students are able to :

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1. Know the relation between central angle and inscribed angle, and cyclic quadrilateral
2. Understand the properties of central angle, inscribed angle, and cyclic quadrilateral
3. Use the properties of central angle, inscribed angle, and cyclic quadrilateral on problem solving process

A central angle is an angle formed by two radii

with the vertex at the center of the circle

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Point O is the center of the circle

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∠AOB is a central angle

Measure of Central angle AOB = measure of intercepted arc AB

�∠AOB = AB = x⁰

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An inscribed angle or angle at the circumference is the angle formed between two chords when they meet on the boundary of the circle.

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Examples : ∠CAB, ∠CDB, ∠ACD, ∠ABD

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Relation of Central Angle and Inscribed Angle

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OA, OB, OC are equal radii

so ΔAOC and Δ BOC are isosceles triangles,

Therefore

∠OCA = ∠OAC

∠OCB = ∠OBC

We can determine that

∠ AOC = 180^o - 2 ∠OAC �∠ BOC = 180^o - 2 ∠OBC

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Look at the circle

∠AOB = 360^o – (∠AOC + ∠BOC)

= 360^o – (180^o – 2 ∠OAC+ 180^o – 2 ∠OBC)

= 360^o – (360^o – 2 ∠OAC+ 180^o – 2 ∠OBC)

= 2 (∠OAC + ∠OBC)

∠AOB = 2 ∠ACB

Conclusion : Angle at the center of a circle is twice of angle at the circumference (central angle is twice of any inscribed angle) subtended by the same arc

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Notice the circle

If ∠ AOB is a straight angle 180⁰ and AB is diameter then ∠ACB = 90⁰ because ∠ AOB and ∠ ACB subtended by the same arc (AB arc)

The measure of angle at semicircle (subtended by the diameter) is 90⁰

Angle at semicircle

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Angles at the circumference (inscribed angles) subtended by the same arc are equal

Notice the circle !

• BAC = ∠ BDC (subtended BC arc)
• ACD = ∠ ABD (subtended AD arc)

Angles at the circumference

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Relation of Central Angle (Major Arc) and Inscribed Angle

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Look at the circle !

Determine the value of x and y

Answer :

Given : ∠ AFB = 40⁰ subtend AB arc

∠ BDC = 25⁰ subtend BC arc

So :

y = ∠ AEC = (40⁰ + 25⁰ ) = 65⁰ inscribed angle subtend AC arc

x = ∠ AOC = 130⁰ (central angle is twice of inscribed angle

subtended by the same arc, AC arc )

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If angle BDC = 32⁰ Find the value of x from the figure !

Answer

∠ BDC = 32⁰

∠ BAC = ∠ BDC = 32⁰ (subtended by BC )

∠ ABC = 90⁰ (inscribed angle subtended by diameter)

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So x = 180⁰ – ∠BAC – ∠ ABC

x = 180 – 32⁰ – 90⁰

x = 58⁰

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Engineering & Architecture

• Road design: Civil engineers calculate the “degree of curvature” for curved roads using central angle to ensure safe vehicle transition
• Mechanical gears: The spacing and size of gear teeth are determined by central angles to achieve specific
• Circular windows and arches: Architects use central angles to define the exact sectors and arc lengths needed for decorative circular features

Navigation & Astronomy

• Great Circle Distance: Navigators use central angles to find the shortest path between two points on Earth’s spherical surface
• Star Trails: In traditional navigation, the arc distance a star travels across the sky is measured as a central angle relative to the observer’s “celestial sphere”
• Astronomy: Scientists calculate the angular distance between celestial bodies to map orbits and predict movements

Technology & Timekeeping: Clock Hands, Time calculation, Minute precision

Data & Design: Pie charts, Ferris Wheels, Food portions

Angles at Circle Real-Life Examples:

Relation of Central Angle and Inscribed Angle

Minor Arc

Major Arc

Semicircle

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Relation of Central Angle and Inscribed Angle

Minor Arc

Major Arc

Cyclic Quadrilateral

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• A cyclic quadrilateral is a quadrilateral with its 4 vertices on the circumference of a circle. This circle is called the circumcircle or circumscribed circle.
• A cyclic quadrilateral is also called chordal quadrilateral, because the sides of the quadrilateral are chords of the circumcircle.

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#1 :

Its opposite angles are supplementary

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∠BAD + ∠BCD = 180^o

∠ABC + ∠ADC = 180^o

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The Other Proof :

∠ BAD = ½ ∠ BOD₁ (green)

∠ BCD = ½ ∠ BOD₂ (red)

• BAD + ∠ BCD = ½ (∠ BOD₁ + ∠ BOD₂)
• BAD + ∠ BCD = ½ (360^o)
• BAD + ∠ BCD = 180^o

Do the identical steps for ∠ ABC and ∠ ADC, then we will get :

∠BAD + ∠BCD = 180^o

∠ABC + ∠ADC = 180^o

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∠ADE = ∠ABC

∠ABC + ∠ADC = 180^o

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∠ADE + ∠ADC = 180^o

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• An exterior angle of a cyclic quadrilateral is equal to the corresponding interior opposite angle.

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Study the picture and then find the magnitude of ∠ADC and ∠BCD!

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Answer :

i) ∠ADC + ∠ABC = 180^o

∠ADC + 95^o = 180^o

∠ADC = 180^o – 95^o = 85^o

ii) ∠BCD + ∠BAC = 180^o

∠BCD + 88^o = 180^o

∠BCD = 180^o – 88^o = 92^o

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Study the picture and then find the magnitude of ∠ADC, ∠BCD, ∠BAD, ∠ABC!

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∠ADC + ∠ABC = 180^o

∠BCD = 180^o – ∠AOB

= 180^o – 97.5^o = 82.5^o

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∠BCD + ∠BAD = 180^o

∠BCD = 180^o – ∠AOB

= 180^o – 79.5^o = 100.5^o

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∠BAD = ½ ∠BOD

= ½ (∠BOC + ∠COD)

= ½ (67^o + 92^o)

= ½ (159^o) = 79.5^o

∠ABC = ½ ∠AOC

= ½ (∠AOD + ∠DOC)

= ½ (103^o + 92^o)

= ½ (195^o) = 97.5^o

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Answer :

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Find the value of x and y

Answer :

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• Architecture & Design: Circular window frames (square, rectangle, or isosceles trapezoid windows) where all four corners touch the circular frame.
• Engineering & Wheels: Bicycle or wagon wheels, where spokes connects the centre to the rim, but specifically, the points of intersection on the rim can form cyclic quadrilaterals.
• Photography/Optics: Camera aperture blades (often 4, 6, or more blades) that close to form a near-circular opening, creating cyclic quadrilateral shapes in the process.
• Arch Bridges: Bridges with circular or semi-circular arches often use isosceles trapezoids (a type of cyclic quadrilateral) for structural support.
• Layout Planning: A rectangular table placed exactly within a circular garden patch, or a rectangular picture frame placed within a circular mat.
• Structural Bracing: In some engineered structures, four points positioned on a circle are connected, providing enhanced stability for roofs or frame designs.

Cyclic Quadrilateral Real-Life Examples:

MATHS PATHWAY BOOK

Practice 9.3 Page 350 – 351

Basic (No. 1 – 8)

Intermediate (No. 9 – 11)

Advanced (No. 12 – 14)

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Link to learn more :

https://www.geogebra.org/m/y89ax4eu

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https://www.geogebra.org/m/r9enz5ny

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https://www.geogebra.org/m/WppgrNHx

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https://www.geogebra.org/m/qh3khzfr

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THANK YOU

MGMP Matematika SMP PENABUR Jakarta