Mathematics questions, chapter 6
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Angles and Tangents of Circles
40 marks. Use Practice to check each answer as you go, or Exam to answer everything first. Show your working and give a reason for each step in Section B. Diagrams are not drawn to scale unless the question says so.
Section A: objective questions
12 marksChoose the correct answer. Each question carries 1 mark.
1
Question 1
Diagram 1 shows a circle with centre O. Which angle is the angle at the circumference subtended by the minor arc PQ?
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B
The angle at the circumference has its vertex on the circle (R) and its arms end at P and Q, the two ends of the arc. ∠POQ has its vertex at the centre, so it is the central angle. ∠OPQ and ∠RPQ have their vertex at P.
2
Question 2
In Diagram 2, P, Q, R and S lie on a circle. ∠PRQ = 38°. Find the value of x.
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C
x and 38° are both angles at the circumference subtended by arc PQ, so they are equal: x = 38°. 76° is the central angle on arc PQ and 19° is half of 38°.
3
Question 3
In Diagram 3, arc PQ = 4 cm and arc RS = 12 cm. Find the value of x.
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A
The angle at the circumference is proportional to the arc length. Arc RS is 12 ÷ 4 = 3 times arc PQ, so x = 3 × 15° = 45°. 60° comes from multiplying by 4 instead of 3; 23° from adding (15 + 12 − 4).
4
Question 4
Diagram 4 shows a circle with centre O. ∠POQ = 112° and R lies on the minor arc PQ. Find the value of x.
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D
R is on the minor arc, so ∠PRQ is subtended by the major arc, whose central angle is 360° − 112° = 248°. x = ½ × 248° = 124°. 56° would be correct only if R were on the major arc.
5
Question 5
In Diagram 5, PQ is a diameter of the circle with centre O. ∠RPQ = 35°. Find the value of x.
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B
The angle in a semicircle ∠PRQ = 90°. In triangle PQR, x = 180° − 90° − 35° = 55°. 90° is ∠PRQ and 125° is 180° − 55°.
6
Question 6
Which statement describes a cyclic quadrilateral?
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A
A cyclic quadrilateral has all four vertices on the circumference. A circle inside that touches the four sides is a different idea and does not make the quadrilateral cyclic.
7
Question 7
In Diagram 6, PQRS is a cyclic quadrilateral and QRT is a straight line. ∠SRT = 82°. Find ∠SPQ.
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D
The exterior angle SRT equals the opposite interior angle SPQ, so ∠SPQ = 82°. Check: ∠QRS = 180° − 82° = 98°, and ∠SPQ + ∠QRS = 180°. 98° is the interior angle at R, not at P.
8
Question 8
Diagram 7 shows a circle with centre O and four straight lines. Which line is a tangent to the circle?
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C
A tangent touches the circle at exactly one point. CD touches once. AB and GH cut the circle at two points, and EF does not touch it at all.
9
Question 9
In Diagram 8, O is the centre of the circle and AB is a tangent that touches the circle at B. ∠AOB = 64°. Find the value of x.
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B
The radius OB is perpendicular to the tangent, so ∠OBA = 90°. x = 180° − 90° − 64° = 26°.
10
Question 10
In Diagram 9, PQR is a tangent to the circle at Q. ∠PQS = 60° and ∠SQT = 72°. Find the value of x.
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A
On the straight line PQR: ∠RQT = 180° − 60° − 72° = 48°. x = ∠QST is in the alternate segment of chord QT, so x = ∠RQT = 48°. 60° is ∠PQS, which matches the angle at T, not at S.
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Question 11
Diagram 10 shows a cyclic quadrilateral PQRS. ∠PQR = (2x + 10)° and ∠PSR = (3x − 20)°. Find the value of x.
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C
∠PQR and ∠PSR are opposite angles, so (2x + 10) + (3x − 20) = 180. 5x − 10 = 180, 5x = 190, x = 38. Check: 86° + 94° = 180°.
12
Question 12
In Diagram 11, PA and PB are tangents to the circle with centre O. ∠APB = 48°. Find the value of x.
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D
PA = PB, so triangle PAB is isosceles and ∠PAB = (180° − 48°) ÷ 2 = 66°. OA is perpendicular to the tangent, so ∠OAP = 90° and x = 90° − 66° = 24°. Check: ∠AOB = 360° − 90° − 90° − 48° = 132°, and (180° − 132°) ÷ 2 = 24°.
Section B: structured questions
28 marksAnswer all questions. Show your working and give a reason for each step.
13
Question 13
Diagram 12 shows a circle with centre O. R lies on the major arc PQ and S lies on the minor arc PQ. ∠POQ = 84°.
[1 mark]
(a)Find ∠PRQ. Give a reason.
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Tick each point you got.
[2 marks]
(b)Find ∠PSQ. Give a reason.
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Check: ∠PRQ + ∠PSQ = 42° + 138° = 180°.
Tick each point you got.
14
Question 14
Diagram 13 shows a circle. Arc PQ = 5 cm and arc QS = 15 cm. ∠PRQ = 24°.
[1 mark]
(a)Find the angle at the centre subtended by arc PQ.
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[2 marks]
(b)Find the value of x.
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[1 mark]
(c)Calculate the circumference of the circle.
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360° ÷ 48° = 7.5, so the whole circumference is 7.5 pieces of 5 cm.
Tick each point you got.
15
Question 15
Diagram 15 shows a cyclic quadrilateral ABCD in a circle with centre O. ∠AOC = 150° and ∠CAD = 40°.
[1 mark]
(a)Find ∠ABC.
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[1 mark]
(b)Find ∠ADC.
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[1 mark]
(c)Find ∠ACD.
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[1 mark]
(d)Find ∠ABD. Give a reason.
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16
Question 16
Diagram 16 shows a circle with centre O and radius 5 cm. PA and PB are tangents to the circle at A and B. OP = 13 cm.
[1 mark]
(a)Calculate the length of PA.
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[2 marks]
(b)Calculate ∠APO, correct to one decimal place.
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[1 mark]
(c)Hence find ∠APB.
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17
Question 17
Diagram 17 shows a circle. PAQ is a tangent to the circle at A. AB = AC and ∠PAB = 66°.
[1 mark]
(a)Find ∠ACB. Give a reason.
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[1 mark]
(b)Find ∠ABC.
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[1 mark]
(c)Find ∠BAC.
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[1 mark]
(d)Show that PQ is parallel to BC.
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18
Question 18
Diagram 18 shows two circles with centres A and B. Their radii are 8 cm and 3 cm. AB = 13 cm. PQ is a common tangent that touches the circles at P and Q.
[1 mark]
(a)State the number of common tangents the two circles have.
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[2 marks]
(b)Calculate the length of PQ.
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[1 mark]
(c)The small circle is moved along AB until it touches the large circle from outside. State the number of common tangents now.
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19
Question 19
In Diagram 19, O is the centre of the circle and ∠AOC = 140°. Aini says, "∠ABC = 70° because the angle at the circumference is half of the central angle."
[1 mark]
(a)Is Aini correct for the point B in Diagram 19?
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No
[2 marks]
(b)Explain Aini's mistake and find the correct value of ∠ABC.
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Aini is right only when B is on the major arc AC. Check with a cyclic quadrilateral: 70° + 110° = 180°.
Tick each point you got.
20
Question 20
O is the centre of a circle. Chord PQ subtends ∠POQ = 100° at the centre. Sketch the circle and mark a point R on the circumference so that ∠PRQ = 50°, and a point S on the circumference so that ∠PSQ = 130°.
[2 marks]
(a)State on which arc (major or minor) PQ each point lies and give a reason.
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R on the major arc PQ (50° = ½ of 100°); S on the minor arc PQ (130° = ½ of 260°). Any point on the correct arc will do.
Tick each point you got.