In triangle ABC, angle A and the side a opposite it are known, and angle B is known. Which equation correctly applies the sine rule to find the side b opposite angle B?
A) a/sinA = b/sinB
B) a/sinB = b/sinA
C) a x sinA = b x sinB
D) sinA/a = sinB x b
(Total for Question 1 is 1 mark)
2
In triangle ABC, angle A = 50 degrees, angle B = 70 degrees and BC = 6 cm (BC is opposite angle A). Calculate the length of AC (opposite angle B). Give your answer correct to 3 significant figures.
Diagram NOT accurately drawn
(Total for Question 2 is 2 marks)
3
In triangle PQR, angle P = 48 degrees, angle Q = 63 degrees and QR = 9.5 cm (QR is opposite angle P). Calculate the length of PR (opposite angle Q). Give your answer correct to 3 significant figures.
Diagram NOT accurately drawn
(Total for Question 3 is 3 marks)
4
In triangle XYZ, XZ = 13 cm (opposite angle Y), XY = 10 cm (opposite angle Z), and angle X = 100 degrees. Calculate angle Z. Give your answer correct to 1 decimal place.
Diagram NOT accurately drawn
(Total for Question 4 is 3 marks)
5
In triangle ABC, angle A = 35 degrees, angle B = 82 degrees and BC = 6.3 cm (BC is opposite angle A). Calculate the length of AB (opposite angle C). Give your answer correct to 3 significant figures.
Diagram NOT accurately drawn
(Total for Question 5 is 4 marks)
6
In triangle ABC, angle BAC = 29 degrees, BC = 14 cm (opposite angle A) and AB = 8 cm (opposite angle C).
Diagram NOT accurately drawn
(a)Calculate the size of angle ACB. Give your answer correct to 1 decimal place.(3)
(b)Hence calculate the length of AC. Give your answer correct to 3 significant figures.(3)
(Total for Question 6 is 6 marks)
7
In triangle ABC, angle A = 30 degrees, angle C = 45 degrees, and side AB = 6*√2 cm (AB is opposite angle C). Show that the length of BC (opposite angle A) is exactly 6 cm.
Diagram NOT accurately drawn
(Total for Question 7 is 3 marks)
8
A ship sails from a lighthouse L to a buoy B, a distance of 12 km, on a bearing of 065 degrees. A harbour H is on a bearing of 116 degrees from L, and on a bearing of 150 degrees from B.
Diagram NOT accurately drawn
(a)Show that angle BLH = 51 degrees.(1)
(b)Show that angle LBH = 95 degrees.(1)
(c)Hence calculate the distance BH, giving your answer correct to 3 significant figures.(4)
(Total for Question 8 is 6 marks)
9
A radio mast is supported by a cable attached to the top, T, of the mast and anchored to the ground at point A. A second cable runs from T to a point B on the ground, on the same side as A, with A, B and the foot of the mast in view. Angle TAB = 62 degrees, angle TBA = 71 degrees, and AB = 45 m. Calculate the length of cable TA. Give your answer correct to 3 significant figures.
Diagram NOT accurately drawn
(Total for Question 9 is 4 marks)
10
The diagram shows quadrilateral ABCD, made up of triangle ABC and triangle ACD joined along the diagonal AC. In triangle ABC, angle BAC = 32 degrees, angle ABC = 110 degrees and BC = 7 cm (opposite angle A). In triangle ACD, angle ACD = 40 degrees and angle ADC = 95 degrees.
Diagram NOT accurately drawn
(a)Calculate the length of AC. Give your answer correct to 3 significant figures.(3)
(b)Hence calculate the length of CD. Give your answer correct to 3 significant figures.(4)
(Total for Question 10 is 7 marks)
11
In triangle ABC, angle A and sides a and b are given (side a is opposite angle A). Which condition guarantees that the ambiguous case (two possible triangles) cannot occur?
A) angle A is acute and a < b
B) angle A is acute and a ≥ b
C) angle A is obtuse and a < b
D) angle A = 90 degrees and a < b
(Total for Question 11 is 1 mark)
12
In triangle ABC, angle A = 35 degrees, BC = 6 cm (opposite angle A) and AC = 9 cm (opposite angle B).
Diagram NOT accurately drawn
(a)Calculate the two possible values of angle B. Give each answer correct to 1 decimal place.(3)
(b)Explain why there are two possible values for angle B.(1)
(Total for Question 12 is 4 marks)
13
In triangle ABC, angle A = 40 degrees, angle B = 65 degrees, BC = (x + 3) cm (opposite angle A) and AC = 2x cm (opposite angle B). Find the value of x. Give your answer correct to 3 significant figures.
Diagram NOT accurately drawn
(Total for Question 13 is 4 marks)
14
From a point A on level ground, the angle of elevation of the top, D, of a tower is 28 degrees. From a point B, 50 m closer to the tower along the same straight line as A and the foot of the tower, the angle of elevation of D is 41 degrees.
Diagram NOT accurately drawn
(a)Show that angle ADB = 13 degrees.(1)
(b)Calculate the length of BD. Give your answer correct to 3 significant figures.(3)
(c)Hence calculate the height of the tower, CD. Give your answer correct to 3 significant figures.(2)
(Total for Question 14 is 6 marks)
15
Two straight roads cross at a junction J, at an angle of 76 degrees. A cafe C is on one road, 850 m from J. A school S is on the other road, with angle JCS = 32 degrees.
Diagram NOT accurately drawn
(a)Calculate the distance CS between the cafe and the school. Give your answer correct to 3 significant figures.(4)
(b)Hence find the shortest distance from the school to the road JC. Give your answer correct to 3 significant figures.(2)
(Total for Question 15 is 6 marks)
16
Triangle ABC has angle C not equal to 90 degrees. The perpendicular from C meets line AB (extended if necessary) at N, with CN = h. Prove that a/sinA = b/sinB, where a = BC and b = AC.
Diagram NOT accurately drawn
(Total for Question 16 is 4 marks)
17
In triangle DEF, angle D = 42 degrees, EF = 68 m (opposite angle D) and DE = 95 m (opposite angle F).
Diagram NOT accurately drawn
(a)Calculate the two possible values of angle F. Give each answer correct to 1 decimal place.(3)
(b)Given that angle DFE is obtuse, calculate the length of DF. Give your answer correct to 3 significant figures.(4)
(c)Explain why, without being told that angle DFE is obtuse, there would be two possible lengths for DF.(1)
(Total for Question 17 is 8 marks)
18
The diagram shows quadrilateral PQRS, made up of triangle PQR and triangle PRS joined along the diagonal PR. In triangle PQR, angle QPR = 34 degrees, angle PQR = 98 degrees and PQ = 6.5 cm (opposite angle R). In triangle PRS, angle PRS = 40 degrees and PS = 10 cm (opposite angle R).
Diagram NOT accurately drawn
(a)Calculate the length of PR. Give your answer correct to 3 significant figures.(3)
(b)Calculate the size of angle PSR. Give your answer correct to 1 decimal place.(3)
(c)Hence calculate the length of RS. Give your answer correct to 3 significant figures.(3)
(Total for Question 18 is 9 marks)
Mark scheme · 7.12 The Sine Rule
Question 1
B1 A oe
Answer: A) a/sinA = b/sinB
Question 2
M1 AC/sin70 = 6/sin50 oe
A1 7.36 cm awrt
Answer: 7.36 cm (3 sf)
Question 3
M1 PR/sin63 = 9.5/sin48 oe
M1 PR = 9.5 x sin63 / sin48
A1 11.4 cm awrt
Answer: 11.4 cm (3 sf)
Question 4
M1 YZ2 = 132 + 102 - 2(13)(10)cos100 oe (cosine rule for side opposite the given angle)
M1 YZ = 17.7 (cm) so sinZ/10 = sin100/17.7 oe
A1 33.8 degrees awrt
Answer: 33.8 degrees (1 dp)
Question 5
M1 angle C = 180 - 35 - 82 (=63 degrees)
M1 AB/sin63 = 6.3/sin35 oe
M1 AB = 6.3 x sin63 / sin35
A1 9.79 cm awrt
Answer: 9.79 cm (3 sf)
Question 6
(a) M1 sin(ACB)/8 = sin29/14 oe
(a) M1 sin(ACB) = 8 x sin29 / 14
(a) A1 16.1 degrees awrt (rejecting the obtuse solution as angle A + angle C would then exceed 180 degrees)
M1 exact values sin30 = 1/2 and sin45 = √2/2 used, BC = 6sqrt(2) x (1/2) / (√2/2)
A1 BC = 6 cm cso (all steps shown with exact values, answer given)
Answer: 6 cm (answer given)
Question 8
(a) B1 116 - 65 = 51 degrees cso
(a) Answer: 51 degrees (answer given)
(b) B1 bearing of L from B = 065+180 = 245, then 245 - 150 = 95 degrees cso
(b) Answer: 95 degrees (answer given)
(c) M1 angle LHB = 180 - 51 - 95 (=34 degrees)
(c) M1 BH/sin51 = 12/sin34 oe
(c) M1 BH = 12 x sin51 / sin34
(c) A1 16.7 km awrt
(c) Answer: 16.7 km (3 sf)
Question 9
B1 angle ATB = 180 - 62 - 71 (=47 degrees)
M1 TA/sin71 = 45/sin47 oe
M1 TA = 45 x sin71 / sin47
A1 58.2 m awrt
Answer: 58.2 m (3 sf)
Question 10
(a) M1 AC/sin110 = 7/sin32 oe
(a) M1 AC = 7 x sin110 / sin32
(a) A1 12.4 cm awrt
(a) Answer: 12.4 cm (3 sf)
(b) B1 angle CAD = 180 - 40 - 95 (=45 degrees)
(b) M1 CD/sin45 = their AC/sin95 oe (ft)
(b) M1 CD = their AC x sin45 / sin95
(b) A1 8.81 cm awrt (ft their AC)
(b) Answer: 8.81 cm (3 sf)
Question 11
B1 B oe
Answer: B) angle A is acute and a ≥ b
Question 12
(a) M1 sinB/9 = sin35/6 oe
(a) A1 59.4 degrees awrt (first solution)
(a) A1 120.6 degrees awrt (180 - first solution)
(a) Answer: 59.4 degrees or 120.6 degrees
(b) C1 because angle A is acute and the side opposite it (BC = 6) is shorter than the other given side (AC = 9), both an acute and an obtuse angle satisfy the same sine value, and both give a valid triangle (angle sum under 180 degrees)
(b) Answer: Because angle A is acute and BC < AC, both the acute and obtuse angles with the same sine value produce a valid triangle (the ambiguous case).
Question 13
M1 (x+3)/sin40 = 2x/sin65 oe
M1 (x+3) x sin65 = 2x x sin40, i.e. 0.9063(x+3) = 1.2856x oe (values seen)
M1 correctly collects x terms, e.g. 2.719 = 0.3793x oe
A1 7.17 awrt
Answer: x = 7.17 (3 sf)
Question 14
(a) B1 41 - 28 = 13 degrees cso (exterior angle of triangle ABD equals the sum of the two opposite interior angles)
(a) Answer: 13 degrees (answer given)
(b) M1 BD/sin28 = 50/sin13 oe
(b) M1 BD = 50 x sin28 / sin13
(b) A1 104 m awrt
(b) Answer: 104 m (3 sf)
(c) M1 CD = their BD x sin41 (using right-angled triangle BCD) ft
(c) A1 68.5 m awrt (ft their BD)
(c) Answer: 68.5 m (3 sf)
Question 15
(a) B1 angle JSC = 180 - 76 - 32 (=72 degrees)
(a) M1 CS/sin76 = 850/sin72 oe
(a) M1 CS = 850 x sin76 / sin72
(a) A1 867 m awrt
(a) Answer: 867 m (3 sf)
(b) M1 shortest distance = their CS x sin32 (perpendicular from S to line JC) ft
(b) A1 460 m awrt (ft their CS)
(b) Answer: 460 m (3 sf)
Question 16
M1 in right triangle ANC, h = b x sinA
M1 in right triangle BNC, h = a x sinB
M1 equates both expressions for h: b sinA = a sinB
A1 divides both sides by sinA sinB to reach a/sinA = b/sinB cso
Answer: a/sinA = b/sinB (proven)
Question 17
(a) M1 sinF/95 = sin42/68 oe
(a) A1 69.2 degrees awrt (first solution)
(a) A1 110.8 degrees awrt (180 - first solution)
(a) Answer: 69.2 degrees or 110.8 degrees
(b) M1 selects F = 110.8 and finds angle E = 180 - 42 - 110.8 (=27.2 degrees)
(b) M1 DF/sin27.2 = 68/sin42 oe
(b) M1 DF = 68 x sin27.2 / sin42
(b) A1 46.5 m awrt
(b) Answer: 46.5 m (3 sf)
(c) C1 because both values of angle F found in part (a) give a valid triangle (angle sums are both less than 180 degrees), each leading to a different value of angle E and hence a different length for DF via the sine rule
(c) Answer: Because both values of angle F give a valid triangle, each produces a different angle E, and so a different value of DF.
Question 18
(a) B1 angle PRQ = 180 - 34 - 98 (=48 degrees)
(a) M1 PR/sin98 = 6.5/sin48 oe
(a) A1 8.66 cm awrt
(a) Answer: 8.66 cm (3 sf)
(b) M1 sin(PSR)/their PR = sin40/10 oe (ft their PR)
(b) A1 33.8 degrees awrt (ft their PR)
(b) B1 rejects the obtuse solution (146.2) since angle P would then be negative (180-40-146.2 < 0), which is impossible