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Topics · Angles in Circles

Properties of circles in DSE Maths Paper 2

This topic is geometry inside a circle. Questions give you a figure with points on a circle and ask for an angle, using the facts about angles at the centre and at the circumference, angles in the same segment, cyclic quadrilaterals, and equal chords and arcs.

What this topic covers

  • Angle at the centre is twice the angle at the circumference
  • Angles in the same segment; the angle in a semicircle
  • Opposite angles of a cyclic quadrilateral
  • Equal chords, equal arcs and equal angles

Two sample questions

Sample 1

In the figure, ABCD is a cyclic quadrilateral and AB is produced to E. If angle DAC equals 72 degrees, angle ACD equals 40 degrees and angle BEC equals 59 degrees, then angle BCE equals

ABCDE
  1. A13 degrees .
  2. B31 degrees .
  3. C53 degrees .
  4. D121 degrees .
Show the answer and solution

Answer: C

Worked solution

∠ADC\displaystyle \angle ADC=180∘\displaystyle {}= 180^\circ−72∘\displaystyle {}- 72^\circ−40∘\displaystyle {}- 40^\circ=68∘\displaystyle {}= 68^\circ \quad(∠\displaystyle (\angle sum of △ACD)\displaystyle \triangle ACD)∠CBE=∠ADC=68∘(\displaystyle \angle CBE = \angle ADC = 68^\circ \quad (ext. ∠\displaystyle \angle of cyclic quad.)\displaystyle )∠BCE\displaystyle \angle BCE=180∘\displaystyle {}= 180^\circ−68∘\displaystyle {}- 68^\circ−59∘\displaystyle {}- 59^\circ=53∘\displaystyle {}= 53^\circ \quad(∠\displaystyle (\angle sum of △BCE)\displaystyle \triangle BCE)

Sample 2

In the figure, O is the centre of the circle ABC. If angle OAB equals 12 degrees and angle OCB equals 37 degrees, then angle OAC equals

12°37°ABCO
  1. A16 degrees .
  2. B41 degrees .
  3. C82 degrees .
  4. D131 degrees .
Show the answer and solution

Answer: B

Worked solution

OA=OB=OC\displaystyle OA = OB = OC∠OBA\displaystyle \angle OBA=∠OAB\displaystyle {}= \angle OAB=12∘,  \displaystyle {}= 12^\circ, \;∠OBC\displaystyle \angle OBC=∠OCB\displaystyle {}= \angle OCB=37∘\displaystyle {}= 37^\circ \quad(\displaystyle (base ∠\displaystyle \angles, isos. △)\displaystyle \triangle)∠ABC=12∘+37∘=49∘\displaystyle \angle ABC = 12^\circ + 37^\circ = 49^\circ∠AOC\displaystyle \angle AOC=2∠ABC\displaystyle {}= 2\angle ABC=98∘\displaystyle {}= 98^\circ \quad(∠\displaystyle (\angle at centre=2×∠\displaystyle = 2 \times \angle at circumference)\displaystyle )∠OAC\displaystyle \angle OAC=12(180∘−98∘)\displaystyle {}= \tfrac{1}{2}(180^\circ - 98^\circ)=41∘\displaystyle {}= 41^\circ \quad(\displaystyle (base ∠\displaystyle \angles, isos. △)\displaystyle \triangle)

Try a whole paper

The free mock is a full 45-question Paper 2 under exam timing, marked the moment you submit, with a report by topic.

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Original material by Dsetopia, written from the EDB syllabus learning objectives. Not affiliated with the HKEAA; no exam content is reproduced.