The diagram shows a circle, with centre OO, passing through AA, BB, CC, DD and EE.
CGCG and FGFG are tangents to the circle at points CC and AA respectively. Angle OAB=56∘OAB = 56^\circ, angle BDE=78∘BDE = 78^\circ and angle OGC=19∘OGC = 19^\circ.
Find, giving reasons for each answer, the value of

  1. (a)

    angle EAFEAF,

    Answer ...

  2. (b)

    angle BOGBOG.

    Answer ...

  3. (23)
    1. (b)

      The game is now programmed such that no two same icons will appear in the display.

      For example, if an interlocking-circles icon is chosen in set 1, this icon will be removed from set 2 and an icon will be randomly chosen from the remaining six icons in set 2 as shown in the diagram below.

      Set 2: [Two interlocking-circles icons crossed out, followed by two sailboat icons, three rocket icons, and one graduate icon]

      If now a graduate icon is chosen in set 2, the display would be [interlocking-circles icon] [graduate icon].

      Draw a tree diagram to show the probabilities of all the possible outcomes.

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Worked solution

22(a)

1. Find angle AOB using the isosceles triangle formed by the radii.

Since OAOA and OBOB are radii, OA=OBOA = OB, making △OAB\triangle OAB an isosceles triangle.
∠OBA=∠OAB=56∘\angle OBA = \angle OAB = 56^\circ
∠AOB=180∘−2(56∘)=68∘\angle AOB = 180^\circ - 2(56^\circ) = 68^\circ (sum of angles in a triangle)

2. Find angle ADB using the angle at centre theorem.

∠ADB=12∠AOB=12(68∘)=34∘\angle ADB = \frac{1}{2} \angle AOB = \frac{1}{2}(68^\circ) = 34^\circ (angle at centre is twice angle at circumference)

3. Find angle ADE and use the alternate segment theorem to find angle EAF.

∠ADE=∠BDE−∠ADB=78∘−34∘=44∘\angle ADE = \angle BDE - \angle ADB = 78^\circ - 34^\circ = 44^\circ
By the alternate segment theorem, the angle between the tangent FAFA and chord AEAE is equal to the angle in the alternate segment.
∠EAF=∠ADE=44∘\angle EAF = \angle ADE = 44^\circ

Answer

44∘44^\circ

22(b)

1. Find angle COG using the tangent–radius right angle

Since CGCG is a tangent and OCOC is a radius, OC⊥CGOC \perp CG, so ∠OCG=90∘\angle OCG = 90^\circ (tangent ⊥\perp radius).

In △OCG\triangle OCG:
∠COG=180∘−90∘−19∘=71∘(angle sum of triangle).\angle COG = 180^\circ - 90^\circ - 19^\circ = 71^\circ \quad (\text{angle sum of triangle}).

2. Use symmetry of the two tangents to find angle AOG

The tangents from external point GG to the circle are equal (GA=GCGA = GC), and △OAG≡△OCG\triangle OAG \equiv \triangle OCG (RHS). Hence OGOG bisects ∠AOC\angle AOC, so
∠AOG=∠COG=71∘.\angle AOG = \angle COG = 71^\circ.

3. Find angle AOB from isosceles triangle OAB

OA=OBOA = OB (radii), so △OAB\triangle OAB is isosceles and ∠OBA=∠OAB=56∘\angle OBA = \angle OAB = 56^\circ (base angles of isosceles triangle).
∠AOB=180∘−56∘−56∘=68∘.\angle AOB = 180^\circ - 56^\circ - 56^\circ = 68^\circ.

4. Combine to find angle BOG

BB lies on the arc between AA and CC, with OGOG inside ∠AOC\angle AOC. Since ∠AOG=71∘\angle AOG = 71^\circ and ∠AOB=68∘\angle AOB = 68^\circ:
∠BOG=∠AOG−∠AOB=71∘−68∘=3∘.\angle BOG = \angle AOG - \angle AOB = 71^\circ - 68^\circ = 3^\circ.

Answer

∠BOG=3∘\angle BOG = 3^\circ

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