The Sine Rule: Fluency and Exam Drill Worksheet with Answers

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7.12D The Sine Rule: Fluency and Exam Drill

EDEXCEL 1MA1 · Calculator allowed · about 70 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

1

In triangle ABC, side a is opposite angle A and side b is opposite angle B. Write down the sine rule connecting a, angle A, b and angle B.

Figure (to be drawn): General triangle ABC with vertices A, B, C. Side a is opposite vertex A, side b is opposite vertex B, side c is opposite vertex C.

(Total for Question 1 is 1 mark)

2

In triangle ABC, angle A = 39 degrees, angle B = 68 degrees, and BC = 7.4 cm (BC is opposite angle A). Calculate the length of AC (opposite angle B). Give your answer correct to 3 significant figures.

Figure (to be drawn): Triangle ABC with angle A = 39 degrees and angle B = 68 degrees marked at their vertices. Side BC = 7.4 cm is labelled opposite vertex A. AC is the unknown side opposite vertex B.

(Total for Question 2 is 2 marks)

3

In triangle DEF, angle D = 52 degrees, angle F = 61 degrees, and EF = 8.6 cm (EF is opposite angle D). Calculate the length of DE (opposite angle F). Give your answer correct to 3 significant figures.

Figure (to be drawn): Triangle DEF with angle D = 52 degrees and angle F = 61 degrees marked at their vertices. Side EF = 8.6 cm is labelled opposite vertex D. DE is the unknown side opposite vertex F.

(Total for Question 3 is 2 marks)

4

In triangle GHI, angle G = 44 degrees, angle I = 97 degrees, and GH = 5.5 cm (GH is opposite angle I). Calculate the length of HI (opposite angle G). Give your answer correct to 3 significant figures.

Figure (to be drawn): Triangle GHI with angle G = 44 degrees and angle I = 97 degrees marked at their vertices. Side GH = 5.5 cm is labelled opposite vertex I. HI is the unknown side opposite vertex G.

(Total for Question 4 is 2 marks)

5

In triangle JKL, angle J = 58 degrees, angle K = 77 degrees, and JK = 10.2 cm (JK is opposite angle L). Calculate the length of KL (opposite angle J). Give your answer correct to 3 significant figures.

Figure (to be drawn): Triangle JKL with angle J = 58 degrees and angle K = 77 degrees marked at their vertices. Side JK = 10.2 cm is labelled opposite vertex L. KL is the unknown side opposite vertex J.

(Total for Question 5 is 3 marks)

6

In triangle MNP, angle M = 63 degrees, angle P = 39 degrees, and MP = 14.7 cm (MP is opposite angle N). Calculate the length of NP (opposite angle M). Give your answer correct to 3 significant figures.

Figure (to be drawn): Triangle MNP with angle M = 63 degrees and angle P = 39 degrees marked at their vertices. Side MP = 14.7 cm is labelled opposite vertex N. NP is the unknown side opposite vertex M.

(Total for Question 6 is 3 marks)

7

In triangle RST, angle R = 72 degrees, ST = 14 cm (opposite angle R), and RS = 9 cm (opposite angle T). Calculate the size of angle T. Give your answer correct to 1 decimal place.

Figure (to be drawn): Triangle RST with angle R = 72 degrees marked at vertex R. Side ST = 14 cm is labelled opposite vertex R. Side RS = 9 cm is labelled opposite vertex T, the unknown angle.

(Total for Question 7 is 2 marks)

8

In triangle UVW, angle W = 41 degrees, UV = 13 cm (opposite angle W), and VW = 8 cm (opposite angle U). Calculate the size of angle U. Give your answer correct to 1 decimal place.

Figure (to be drawn): Triangle UVW with angle W = 41 degrees marked at vertex W. Side UV = 13 cm is labelled opposite vertex W. Side VW = 8 cm is labelled opposite vertex U, the unknown angle.

(Total for Question 8 is 2 marks)

9

In triangle XYZ, angle Y = 109 degrees, XZ = 19 cm (opposite angle Y), and XY = 8 cm (opposite angle Z). Calculate the size of angle Z. Give your answer correct to 1 decimal place.

Figure (to be drawn): Triangle XYZ with the obtuse angle Y = 109 degrees marked at vertex Y. Side XZ = 19 cm is labelled opposite vertex Y. Side XY = 8 cm is labelled opposite vertex Z, the unknown angle.

(Total for Question 9 is 2 marks)

10

In triangle CDF, angle F = 97 degrees, CD = 16 cm (opposite angle F), and CF = 7 cm (opposite angle D). Calculate the size of angle D. Give your answer correct to 1 decimal place.

Figure (to be drawn): Triangle CDF with the obtuse angle F = 97 degrees marked at vertex F. Side CD = 16 cm is labelled opposite vertex F. Side CF = 7 cm is labelled opposite vertex D, the unknown angle.

(Total for Question 10 is 2 marks)

11

In triangle LMN, angle L = 115 degrees, MN = 18 cm (opposite angle L), and LN = 9 cm (opposite angle M). Calculate the size of angle N. Give your answer correct to 1 decimal place.

Figure (to be drawn): Triangle LMN with the obtuse angle L = 115 degrees marked at vertex L. Side MN = 18 cm is labelled opposite vertex L. Side LN = 9 cm is labelled opposite vertex M.

(Total for Question 11 is 3 marks)

12

In triangle STU, angle S = 34 degrees, angle U = 81 degrees, and ST = 22 m (ST is opposite angle U). Calculate the length of TU (opposite angle S). Give your answer correct to 3 significant figures.

Figure (to be drawn): Triangle STU with angle S = 34 degrees and angle U = 81 degrees marked at their vertices. Side ST = 22 m is labelled opposite vertex U. TU is the unknown side opposite vertex S.

(Total for Question 12 is 2 marks)

13

In triangle VWX, angle W = 42 degrees, VX = 10.4 cm (opposite angle W), and WX = 7.8 cm (opposite angle V). Calculate the size of angle V. Give your answer correct to 1 decimal place.

Figure (to be drawn): Triangle VWX with angle W = 42 degrees marked at vertex W. Side VX = 10.4 cm is labelled opposite vertex W. Side WX = 7.8 cm is labelled opposite vertex V, the unknown angle.

(Total for Question 13 is 2 marks)

14

In triangle YZA, angle Y = 96 degrees, ZA = 15.2 cm (opposite angle Y), and YA = 8.6 cm (opposite angle Z). Calculate the size of angle Z. Give your answer correct to 1 decimal place.

Figure (to be drawn): Triangle YZA with the obtuse angle Y = 96 degrees marked at vertex Y. Side ZA = 15.2 cm is labelled opposite vertex Y. Side YA = 8.6 cm is labelled opposite vertex Z, the unknown angle.

(Total for Question 14 is 2 marks)

15

A footpath runs from a car park C to a trig point T on a bearing of 048 degrees, a distance of 3.6 km. A bench B lies on a bearing of 106 degrees from C, and on a bearing of 322 degrees from T.

Figure (to be drawn): Bearings diagram: north lines drawn at C and at T. CT = 3.6 km on a bearing of 048 degrees from C. CB is on a bearing of 106 degrees from C. TB is on a bearing of 322 degrees from T. Triangle CTB is formed by joining B to both C and T.

(a)Show that angle BCT = 58 degrees and angle CTB = 94 degrees.(2)

(b)Hence calculate the distance TB from the trig point to the bench. Give your answer correct to 3 significant figures.(3)

(Total for Question 15 is 5 marks)

16

Two coastguard stations, P and Q, are 8 km apart along a straight stretch of coastline. A distress flare is spotted out at sea, at point F. The angle FPQ = 63 degrees and the angle FQP = 79 degrees.

Figure (to be drawn): Coastline PQ = 8 km drawn horizontally. F is out at sea, above the line. angle FPQ = 63 degrees at P, angle FQP = 79 degrees at Q. Triangle PQF is formed.

(a)Calculate the size of angle PFQ.(1)

(b)Calculate the distance QF, from station Q to the flare. Give your answer correct to 3 significant figures.(3)

(c)Hence calculate the shortest (perpendicular) distance from the flare to the coastline PQ. Give your answer correct to 3 significant figures.(3)

(Total for Question 16 is 7 marks)

17

The diagram shows quadrilateral WXYZ, made up of triangle WXY and triangle WYZ joined along the diagonal WY. In triangle WXY, angle XWY = 51 degrees, angle WXY = 72 degrees, and WX = 9.4 cm. In triangle WYZ, angle ZWY = 34 degrees and angle WZY = 95 degrees.

Figure (to be drawn): Quadrilateral WXYZ split by diagonal WY into triangle WXY (angle XWY = 51 degrees at W, angle WXY = 72 degrees at X, WX = 9.4 cm) and triangle WYZ (angle ZWY = 34 degrees at W, angle WZY = 95 degrees at Z).

(a)Calculate the length of the diagonal WY. Give your answer correct to 3 significant figures.(3)

(b)Hence calculate the length of YZ. Give your answer correct to 3 significant figures.(3)

(Total for Question 17 is 6 marks)

18

In triangle ABC, angle A = 35 degrees, BC = 6 cm (opposite angle A), and AC = 9 cm (opposite angle B). This information gives two possible triangles.

Figure (to be drawn): Triangle ABC with angle A = 35 degrees marked at vertex A. Side BC = 6 cm is opposite vertex A. Side AC = 9 cm is opposite vertex B. Two possible positions of B are indicated, showing the ambiguous case.

(a)Calculate the two possible values of angle B. Give your answers correct to 1 decimal place.(2)

(b)Hence find the two corresponding possible values of angle C.(2)

(c)Explain why this information gives two possible triangles.(1)

(Total for Question 18 is 5 marks)

19

In triangle ABC, angle A = 50 degrees, angle B = 65 degrees, BC = (2x + 1) cm (opposite angle A), and AC = (3x - 2) cm (opposite angle B). Use the sine rule to find the value of x. Give your answer correct to 2 decimal places.

Figure (to be drawn): Triangle ABC with angle A = 50 degrees and angle B = 65 degrees marked at their vertices. Side BC = (2x + 1) cm is labelled opposite vertex A. Side AC = (3x - 2) cm is labelled opposite vertex B.

(Total for Question 19 is 4 marks)

20

In triangle LNP, angle L = 60 degrees, angle N = 45 degrees, and LP = 12 cm (LP is opposite angle N). Show that the length of NP (opposite angle L) is exactly 6*√6 cm.

Figure (to be drawn): Triangle LNP with angle L = 60 degrees and angle N = 45 degrees marked at their vertices. Side LP = 12 cm is labelled opposite vertex N. NP is the unknown side opposite vertex L.

(Total for Question 20 is 3 marks)

21

Points A and B lie on level ground, 60 m apart, with the foot of a tall building on the same straight line as A and B, and B closer to the building. The angle of elevation of the top, T, of the building is 33 degrees from A and 51 degrees from B.

Figure (to be drawn): Points A and B on a horizontal line, B between A and the foot of the building. T is the top of the building. Angle of elevation from A to T is 33 degrees; angle of elevation from B to T is 51 degrees. Triangle ABT has angle TAB = 33 degrees at A and angle ABT = 129 degrees at B (the supplement of the 51 degree elevation, since T is on the far side of B from A).

(a)Show that angle ATB = 18 degrees.(3)

(b)Calculate the length BT. Give your answer correct to 3 significant figures.(3)

(c)Hence calculate the height of the building. Give your answer correct to 3 significant figures.(2)

(Total for Question 21 is 8 marks)

22

In triangle ABC, angle A = 40 degrees, angle C = 95 degrees, and the perimeter of the triangle is 32 cm. Calculate the length of BC (opposite angle A). Give your answer correct to 3 significant figures.

Figure (to be drawn): Triangle ABC with angle A = 40 degrees and angle C = 95 degrees marked at their vertices. The perimeter of the triangle is 32 cm. BC, opposite angle A, is the unknown side.

(Total for Question 22 is 4 marks)

Mark scheme · 7.12D The Sine Rule: Fluency and Exam Drill

Question 1

B1 a/sin(A) = b/sin(B) oe (accept sin(A)/a = sin(B)/b)
Answer: a/sin(A) = b/sin(B)

Question 2

M1 AC/sin(68) = 7.4/sin(39) oe
A1 awrt 10.9 (cm)
Answer: AC = 10.9 cm (3 s.f.)

Question 3

M1 DE/sin(61) = 8.6/sin(52) oe
A1 awrt 9.55 (cm)
Answer: DE = 9.55 cm (3 s.f.)

Question 4

M1 HI/sin(44) = 5.5/sin(97) oe
A1 awrt 3.85 (cm)
Answer: HI = 3.85 cm (3 s.f.)

Question 5

M1 angle L = 180 - 58 - 77 = 45 (degrees)
M1 KL/sin(58) = 10.2/sin(45) oe
A1 awrt 12.2 (cm)
Answer: KL = 12.2 cm (3 s.f.)

Question 6

M1 angle N = 180 - 63 - 39 = 78 (degrees)
M1 NP/sin(63) = 14.7/sin(78) oe
A1 awrt 13.4 (cm)
Answer: NP = 13.4 cm (3 s.f.)

Question 7

M1 sin(T)/9 = sin(72)/14 oe
A1 awrt 37.7 (degrees)
Answer: angle T = 37.7 degrees (1 d.p.)

Question 8

M1 sin(U)/8 = sin(41)/13 oe
A1 awrt 23.8 (degrees)
Answer: angle U = 23.8 degrees (1 d.p.)

Question 9

M1 sin(Z)/8 = sin(109)/19 oe
A1 awrt 23.5 (degrees)
Answer: angle Z = 23.5 degrees (1 d.p.)

Question 10

M1 sin(D)/7 = sin(97)/16 oe
A1 awrt 25.7 (degrees)
Answer: angle D = 25.7 degrees (1 d.p.)

Question 11

M1 sin(M)/9 = sin(115)/18 oe
M1 angle M = awrt 26.9 (degrees), dep on correct sine rule setup
A1 angle N = 180 - 115 - their M = awrt 38.1 (degrees) ft
Answer: angle N = 38.1 degrees (1 d.p.)

Question 12

M1 TU/sin(34) = 22/sin(81) oe
A1 awrt 12.5 (m)
Answer: TU = 12.5 m (3 s.f.)

Question 13

M1 sin(V)/7.8 = sin(42)/10.4 oe
A1 awrt 30.1 (degrees)
Answer: angle V = 30.1 degrees (1 d.p.)

Question 14

M1 sin(Z)/8.6 = sin(96)/15.2 oe
A1 awrt 34.2 (degrees)
Answer: angle Z = 34.2 degrees (1 d.p.)

Question 15

(a) B1 angle BCT = 106 - 48 = 58 (degrees), from the difference of the two bearings at C
(a) B1 angle CTB = 322 - (48 + 180) = 94 (degrees), using the back bearing from T to C (048 + 180 = 228) subtracted from 322
(a) Answer: angle BCT = 58 degrees; angle CTB = 94 degrees
(b) M1 angle CBT = 180 - 58 - 94 = 28 (degrees)
(b) M1 TB/sin(58) = 3.6/sin(28) oe
(b) A1 awrt 6.50 (km)
(b) Answer: TB = 6.50 km (3 s.f.)

Question 16

(a) B1 38 (degrees), from 180 - 63 - 79
(a) Answer: angle PFQ = 38 degrees
(b) M1 QF/sin(63) = 8/sin(38) oe
(b) A1 awrt 11.6 (km)
(b) Answer: QF = 11.6 km (3 s.f.)
(c) M1 identify the right-angled triangle formed by dropping a perpendicular from F to line PQ, with hypotenuse QF and angle FQP = 79 at Q
(c) M1 distance = their QF x sin(79) oe (ft their QF)
(c) A1 awrt 11.4 (km)
(c) Answer: perpendicular distance = 11.4 km (3 s.f.)

Question 17

(a) M1 angle WYX = 180 - 51 - 72 = 57 (degrees)
(a) M1 WY/sin(72) = 9.4/sin(57) oe
(a) A1 awrt 10.7 (cm)
(a) Answer: WY = 10.7 cm (3 s.f.)
(b) M1 angle ZYW = 180 - 34 - 95 = 51 (degrees)
(b) M1 YZ/sin(34) = their WY/sin(95) oe (ft their WY)
(b) A1 awrt 5.98 (cm) ft their WY
(b) Answer: YZ = 5.98 cm (3 s.f.)

Question 18

(a) M1 sin(B)/9 = sin(35)/6 oe
(a) A1 awrt 59.4 and awrt 120.6 (degrees), both values required
(a) Answer: angle B = 59.4 degrees or angle B = 120.6 degrees
(b) M1 angle C = 180 - 35 - their B for each case (ft)
(b) A1 awrt 85.6 and awrt 24.4 (degrees), both values required, correctly paired with their B
(b) Answer: angle C = 85.6 degrees (when angle B = 59.4) or angle C = 24.4 degrees (when angle B = 120.6)
(c) B1 correct explanation, e.g. because BC (the side opposite the given angle A) is shorter than AC, so BC can swing to meet the base in two different positions, giving an acute and an obtuse value for angle B that both produce a valid triangle
(c) Answer: BC is shorter than AC, so BC can be drawn in two different positions from C, giving two valid triangles (the ambiguous case).

Question 19

M1 (3x - 2)/sin(65) = (2x + 1)/sin(50) oe
M1 (3x - 2) x sin(50) = (2x + 1) x sin(65) oe, cross-multiplied
M1 correct collection of x terms, e.g. x(3sin50 - 2sin65) = sin65 + 2sin50 oe
A1 awrt 5.02
Answer: x = 5.02 (2 d.p.)

Question 20

M1 NP/sin(60) = 12/sin(45) oe
M1 substitute exact values sin(60) = √3/2 and sin(45) = √2/2
A1 NP = 12 x √3/√2 = 6*√6 cm, correctly simplified (cso)
Answer: NP = 6*√6 cm (exact)

Question 21

(a) M1 angle ABT = 180 - 51 = 129 (degrees), the angle inside triangle ABT at B is the supplement of the elevation angle since T is beyond B
(a) M1 angle ATB = 180 - 33 - 129 oe
(a) A1 = 18 (degrees), correctly shown (cso)
(a) Answer: angle ATB = 18 degrees
(b) M1 BT/sin(33) = 60/sin(18) oe
(b) A1 awrt 106 (m)
(b) Answer: BT = 106 m (3 s.f.)
(c) M1 height = their BT x sin(51) oe (ft their BT), using the right-angled triangle formed by B, the foot of the building, and T
(c) A1 awrt 82.2 (m) ft their BT
(c) Answer: height = 82.2 m (3 s.f.)

Question 22

M1 angle B = 180 - 40 - 95 = 45 (degrees)
M1 each side written as a common multiple k of the sine of its opposite angle, e.g. BC = k sin(40), AC = k sin(45), AB = k sin(95)
M1 k(sin40 + sin45 + sin95) = 32, leading to a value for k
A1 BC = their k x sin(40) = awrt 8.77 (cm)
Answer: BC = 8.77 cm (3 s.f.)

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