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CAT 2025 Lesson : Circles - Angle Properties

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6. Angle Properties

Properties Figure
Property 1:All angles subtended by an arc on the same side of its segment are equal.
∠\angle∠AWB =∠= \angle=∠AXB and ∠\angle∠AZB =∠= \angle=∠AYB

Sum of angles subtended in the major and minor arcs =180o= 180^\mathrm{o}=180o
∠\angle∠AWB +∠+ \angle+∠AYB =180o= 180^\mathrm{o}=180o
Property 2:Angles subtended by the end points of the diameter at any point on the circle is 90o90^\mathrm{o}90o

As XZ is the diameter, ∠\angle∠XAZ =∠= \angle=∠XBZ =∠= \angle=∠XCZ =90o= 90^\mathrm{o}=90o
Property 3:The angle subtended by two points of the circle at the centre is twice the angle subtended by those two points at any other point in the major segment, i.e., Central Angle =2×= 2 \times=2× Inscribed Angle

Where ∠\angle∠AXB =θ= \theta=θ, the central angle ∠\angle∠AOB =2θ= 2 \theta=2θ

Example 6

In the adjoining figure, chord ED is parallel to the diameter AC of the circle. If ∠\angle∠CBE =65o= 65^\mathrm{o}=65o, then what is the value of ∠\angle∠DEC?
[CAT 2004]

(1) 35o35^\mathrm{o}35o                     (2) 55o55^\mathrm{o}55o                     (3) 45o45^\mathrm{o}45o                     (4) 25o25^\mathrm{o}25o                    

Solution

Central angle =2×= 2 \times=2× Inscribed Angle
∴\therefore∴ ∠\angle∠COE =2×= 2 \times=2× ∠\angle∠CBE =130o= 130^\mathrm{o}=130o

As OC === OE === Radius,
△\triangle△EOC is isosceles and ∠\angle∠OEC =∠= \angle=∠OCE =25o= 25^\mathrm{o}=25o

As AC | | ED,
∠\angle∠OCE =∠= \angle=∠OED =25o= 25^\mathrm{o}=25o (Alternate Interior Angles)

Answer: (4) 25o25^\mathrm{o}25o


Example 7

In the figure below, A, B and C lie on the circle with centre O such that O lies on AD, and AC === CD. If ∠\angle∠ACD =140o= 140^\mathrm{o}=140o, then ∠\angle∠ABC = ?

Solution

As △\triangle△ACD is isosceles, ∠\angle∠CDA =∠= \angle=∠CAD = 20o20^\mathrm{o}20o

∠\angle∠ACE =90o= 90^\mathrm{o}=90o (Angle subtended by diameter AE)

In △\triangle△ACE, ∠\angle∠AEC =180−90−20=70o= 180 - 90 - 20 = 70^\mathrm{o}=180−90−20=70o

Angles subtended in major & minor arc are supplementary.
∴\therefore∴ ∠\angle∠ABC +∠+ \angle+∠AEC === 180o180^\mathrm{o}180o
⇒ ∠\angle∠ABC =180o−70o=110o= 180^\mathrm{o} - 70^\mathrm{o} = 110^\mathrm{o}=180o−70o=110o

Answer: 110o110^\mathrm{o}110o


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