ANSWERS. 1. (a) l = rθ or ACB = 2 OA (M1) = 30 cm (A1) (C2) (b) AÔB (obtuse) = 2π 2 (A1) 1. Area = θ r 2 1. = (2π 2)(15) 2 (M1)(A1)

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Eustace Hood
5 years ago
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1 ANSWERS. (a) l = rθ or ACB = OA = 30 cm () (C) (b) AÔB (obtuse) = π () Area = θ r = (π )(5) () = 48 cm (3 sf) () (C [6] 80. AB = rθ = r θ r () = () = 8 cm () OR (5.4) θ =.6 4 θ =.7 (=.48 radians) AB = rθ () 4 = = 8 cm () (C [4]. (a) METHOD choosing cosine rule substituting correctly e.g. AB = (3.9)(3.9) cos.8 AB = 6.(cm) N

2 METHOD evidence of approach involving right-angled triangles substituting correctly x e.g. sin 0.9 =, AB = 3.9 sin AB = 6. (cm) N METHOD 3 choosing the sine rule substituting correctly sin sin.8 e.g. = 3.9 AB AB = 6. (cm) N (b) METHOD reflex AÔB = π.8 (= 4.483) (A) correct substitution A = (3.9) ( ) area = 34. (cm ) N METHOD finding area of circle A = π(3.9) (= ) () finding area of (minor) sector A = (3.9) (.8) (= ) () subtracting M e.g. π(3.9) 0.5(3.9) (.8), area = 34. (cm ) N METHOD 3 finding reflex AÔB = π.8 (= 4.483) (A) finding proportion of total area of circle π.8 θ e.g. π(3.9), πr π π area = 34. (cm) N [7]

3 3. METHOD Evidence of correctly substituting into l = rθ Evidence of correctly substituting into A = For attempting to solve these equations eliminating one variable correctly r θ r = 5 θ =.6 (= 9.7 ) N3 METHOD Setting up and equating ratios 4 80 = πr πr Solving gives r = 5 rθ = 4 or r θ =80 θ =.6 (= 9.7 ) r = 5 θ =.6 (= 9.7 ) N3 [6]

4 4. Perimeter = 5(π ) + 0 () Note: Award for working in radians; () for π ; () for +0. = (0π + 5) cm (= 36.4, to 3 sf) () (C [4] 5. METHOD Evidence of correctly substituting into A = Evidence of correctly substituting into l = rθ For attempting to eliminate one variable leading to a correct equation in one variable r θ π r = 4 θ = (= 0.54, 30 ) N3 6 METHOD Setting up and equating ratios 4 π π 3 = 3 πr πr Solving gives r = 4 rθ = π or r 3 4 θ= π 3 θ = π 6 ( = 0.54,30 ) π r = 4 θ = ( = 0.54,30 ) N3 6 [6]

5 6. (a) For using perimeter = r + r + arc length 0 = r + rθ 0 r θ= r AG N0 ( ) 0 r (b) Finding A = r = 0r r () r For setting up equation in r M Correct simplified equation, or sketch eg 0r r = 5, r 0r + 5 = 0 () r = 5 cm N [6] 7. (a) evidence of appropriate approach M e.g. 3π = π r 9 r =3.5 (cm) N (b) adding two radii plus 3π perimeter = 7+3π (cm) (= 36.4) N (c) evidence of appropriate approach M e.g. 3.5 π 9 area = 0.5π (cm ) (= 63.6) N [6]

6 8. (a) A= r θ 7 = (.5) r r = 36 r = 6cm () () () (C4) (b) Arc length = rθ =.5 6 Arc length = 9 cm () (C) Note: Penalize a total of ( mark) for missing units. [6] 9. (a) correct substitution in l = rθ () π e.g. 0, π π 0π arc length = = 6 3 N (b) area of large sector = π 00π 0 = () 3 6 area of small sector = π 64π 8 = 3 6 () evidence of valid approach (seen anywhere) M e.g. subtracting areas of two sectors, π (0 8 ) 3 36π area shaded = 6π accept, etc. 6 N3 [6]

7 0. OTˆ A = 90 6 () AT = 6 3 = TÔA π = 60 = 3 () Area = area of triangle area of sector π = =.3 cm (or 8 3 6π) () (C4) OR TÔA = 60 () Area of Δ = 6 sin 60 () π Area of sector = () Shaded area = 8 3 6π =.3 cm (3 sf) () (C4) [4]. METHOD Area sector OAB (5) = (0.8) =0 () ( ) ON = 5cos0.8 = () ( ) AN = 5sin 0.8 = () Area of Δ AON = ON AN = (cm ) () Shaded area = = 3.75 (cm ) () (C6)

8 METHOD A O N B F Area sector ABF = (5) (.6) = 0 () Area Δ OAF = (5) sin.6 =.5 () Twice the shaded area = 0.5 ( = 7.5) Shaded area = (7.5) = 3.75 (cm ) () [6] (C6)

9 . h = r so r = 00 r = 50 l = 0θ π = πr 50 θ = 0 () π5 = 0 θ = π = 4.44 (3sf) () (C Note: Accept either answer. [4] 3. (a) choosing sine rule correct substitution AD 4 e.g. = sin0.8 sin 0.3 AD = 9.7 (cm) N (b) METHOD finding angle OAD = π. = (.04) (seen anywhere) choosing cosine rule correct substitution e.g. OD = cos(π.) () OD =. (cm) N3 METHOD finding angle OAD = π. = (.04) (seen anywhere) choosing sine rule correct substitution () e.g. OD sin(π.) = 9.7 sin 0.8 = 4 sin 0.3 OD =. (cm) N3

10 (c) correct substitution into area of a sector formula () e.g. area = area = 6.4 (cm ) N

11 (d) substitution into area of triangle formula OAD correct substitution e.g. A = 4. sin 0.8, A = sin.04, A =. 9.7 sin 0.3 subtracting area of sector OABC from area of triangle OAD e.g. area ABCD = area ABCD = 0.9 (cm ) N [3] 4. (a) Area = r θ = (5 )() = 5 (cm ) () (C) (b) Area OAB = 5 sin = 0.3 () Area = =.7 (cm ) = 3 (3 sf) () (C) [4] 5. (a) area of sector ΑΒDC = π() = π 4 () area of segment BDCP = π area of ΔABC = π () (C3) (b) BP = () area of semicircle of radius BP = π( ) = π () area of shaded region = π (π ) = () (C [6]

12 6. Notes: Candidates may have differing answers due to using approximate answers from previous parts or using answers from the GDC. Some leeway is provided to accommodate this. (a) METHOD Evidence of using the cosine rule eg cos C = a + b c ab, a = b + c bc cos A Correct substitution eg cos AÔP =,4 = cos AÔP 3 cos AÔP = 0.5 AÔP 6π =.8 = (radians) N 45 METHOD Area of AOBP = 5.8 (from part (d)) Area of triangle AOP = = sin AÔP AÔP AÔP =.3 or.8 6π =.8 = (radians) 45 N (b) AÔB = (π.8) (= π 3.64) () 38π =.64 = (radians) N 45 (c) (i) Appropriate method of finding area eg area = θr

13 Area of sector PAEB = 4.63 = 3.0 (cm ) (accept the exact value 3.04) N (ii) Area of sector OADB = 3.64 =.9 (cm ) N (d) (i) Area AOBE = Area PAEB Area AOBP (= ) M = 7.9 (accept 7.3 from the exact answer for PAEB) N (ii) Area shaded = Area OADB Area AOBE (=.9 7.9) M = 4.7 (accept answers between 4.63 and 4.7) N [4] 7. Note: Do not penalize missing units in this question. (a) AB = + = ( cos 75 ) () = ( cos 75 ) AB = ( cos 75 ) (AG) Note: The second () is for transforming the initial expression to any simplified expression from which the given result can be clearly seen. (b) PÔB = 37.5 () BP = tan 37.5 = 9. cm () OR BPˆA = 05 BÂP = 37.5 () AB BP = sin05 sin 37. 5

14 ABsin 37.5 BP = = 9.(cm) () 3 sin 05 (c) (i) Area OBP = 9. or tan = 55.3 (cm ) (accept 55. cm ) () (ii) Area ABP = (9.) sin05 = 4.0 (cm ) (accept 40.9 cm ) () 4 π 75 (d) Area of sector = 75 or π = 94. (cm ) (accept 30π or 94.3 (cm )) ()

15 (e) Shaded area = area OPB area sector = 6.4 (cm ) (accept 6. cm, 6.3 cm ) () [3] 8. (a) appropriate approach e.g. 6 = 8θ AÔC = 0.75 N (b) evidence of substitution into formula for area of triangle e.g. area = 8 8 sin(0.75) area =.8 () evidence of substitution into formula for area of sector e.g. area = area of sector = 4 () evidence of substituting areas e.g. r θ absin C, area of sector area of triangle area of shaded region =.9 cm N4 (c) attempt to set up an equation for area of sector e.g. 45 = 8 θ CÔE =.4065 (.4 to 3 sf) N (d) METHOD attempting to find angle EOF e.g. π EÔF = (seen anywhere) evidence of choosing cosine rule correct substitution e.g. EF = cos EF = 7.57 cm N3

16 METHOD attempting to find angles that are needed e.g. angle EOF and angle OEF EÔF = and OÊF (or OFˆ E) = evidence of choosing sine rule correct substitution () EF 8 e.g. = sin0.985 sin.08 EF = 7.57 cm N3 METHOD 3 attempting to find angle EOF e.g. π EÔF = (seen anywhere) evidence of using half of triangle EOF e.g. x = 8 sin correct calculation e.g. x = 3.78 EF = 7.57 cm N3 [5]

17 9. (a) (i) OP = PQ (= 3cm) R So Δ OPQ is isosceles AG N (ii) Using cos rule correctly eg cos OPˆQ = cos OPˆQ = = 8 8 cos OPˆQ = AG N0 9 (iii) Evidence of using sin A + cos A = M sin OPˆQ = 80 = sin OPˆQ = AG N0 9 (iv) Evidence of using area triangle OPQ = OP PQ sin P M eg , ( ) ( ) Area triangle OPQ = = 0 = 4.47 N (b) (i) OPˆQ = OPˆQ =.46 N (ii) Evidence of using formula for area of a sector eg Area sector OPQ = = 6.57 N π.4594 (c) Q ÔP = ( = 0.84) ()

18 Area sector QOS = = 6.73 N (d) Area of small semi-circle is 4.5π (= ) Evidence of correct approach eg Area = area of semi-circle area sector OPQ area sector QOS + area triangle POQ Correct expression eg 4.5π , 4.5π ( ), M 4.5π ( ) Area of the shaded region = 5.3 N [7] 0. (a) evidence of using area of a triangle e.g. A= sinθ A = sin θ N (b) METHOD PÔA = π θ () area ΔOPA = sin ( π θ) (= sin (π θ)) since sin (π θ) = sin θ R then both triangles have the same area AG N0 METHOD triangle OPA has the same height and the same base as triangle OPB R3 then both triangles have the same area AG N0 ( ) ( ) (c) area semi-circle = π = π area Δ APB = sin θ + sin θ (= 4 sin θ) S = area of semicircle area ΔAPB (= π 4 sin θ) M

19 S = (π sin θ) AG N0 (d) METHOD attempt to differentiate ds e.g. = 4 cos θ dθ setting derivative equal to 0 correct equation e.g. 4 cos θ = 0, cos θ = 0, 4 cos θ = 0 π θ = N3 EITHER evidence of using second derivative Sʹ ʹ (θ) = 4 sin θ π Sʹ ʹ = 4 π it is a minimum because Sʹ ʹ >0 R N0 OR evidence of using first derivative π for θ <, S ʹ (θ) < 0 (may use diagram) π for θ >, S ʹ (θ) > 0 (may use diagram) it is a minimum since the derivative goes from negative to positive R N0 METHOD π 4 sin θ is minimum when 4 sin θ is a maximum R3 4 sin θ is a maximum when sin θ = (A)

20 π θ = A3 N3 (e) S is greatest when 4 sin θ is smallest (or equivalent) (R) θ = 0 (or π) N [8]

OC = $ 3cos. 1 (5.4) 2 θ = (= radians) (M1) θ = 1. Note: Award (M1) for identifying the largest angle.

OC = $ 3cos. 1 (5.4) 2 θ = (= radians) (M1) θ = 1. Note: Award (M1) for identifying the largest angle. 4 + 5 7 cos α 4 5 5 α 0.5. Note: Award for identifying the largest angle. Find other angles first β 44.4 γ 4.0 α 0. (C4) Note: Award (C) if not given to the correct accuracy.. (a) p (C) 4. (a) OA A is