DSE Mathematics Practice Paper: Trigonometry And Coordinate Geometry

DSE Mathematics Practice Paper: Trigonometry and Coordinate Geometry

2026-11-03

DSE Mathematics: Trigonometry and Coordinate Geometry

DSE Mathematics practice paper: trigonometry and coordinate geometry, 12 questions, 100 marks, 75 minutes, covering sine and cosine rules, sectors and circle equations.

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Paper details　Level: S6 | Subject: Mathematics | Time: 75 minutes | Marks: 100 | Questions: 12

Section A: Trigonometric ratios and coordinates (Questions 1-4, 24 marks)

Answer all questions. Show the formula, the substitution and the units, giving answers to 3 significant figures unless stated otherwise.

#	Question	Answer and marking notes
1	In triangle ABC, angle C = 90°, AB = 13 cm and BC = 5 cm. (a) Find AC. (b) Find sin A and tan A as fractions in lowest terms. (c) Find angle A, giving the answer to 3 significant figures. (6 marks)	AC = square root of (13 squared – 5 squared) = 12 cm (2). sin A = BC/AB = 5/13 and tan A = BC/AC = 5/12 (2; lose 1 if only decimals are given). Angle A = inverse tan of 5/12 = 22.6 degrees (2).
2	In triangle ABC, a = 8 cm, b = 11 cm and angle C = 65°. (a) Use the cosine rule to find c. (b) Use the sine rule to find angle A, to 3 significant figures. (6 marks)	c squared = 64 + 121 – 176 cos 65 = 110.6 (2), so c = 10.5 cm (1). sin A = 8 sin 65 / 10.5 = 0.6894 (2), so angle A = 43.6 degrees (1; the obtuse alternative is rejected because b is greater than a).
3	A sector has radius 12 cm and angle 1.8 radians. (a) Find the arc length. (b) Find the area of the sector. (c) Convert 1.8 radians to degrees, to 3 significant figures. (6 marks)	Arc = r times theta = 12 times 1.8 = 21.6 cm (2). Area = half r squared theta = 0.5 times 144 times 1.8 = 129.6 cm squared (2). 1.8 times 180 / pi = 103.1 degrees (2).
4	Given A(-3, 4) and B(5, -2). (a) Find the length of AB. (b) Find the coordinates of the midpoint of AB. (c) Find the equation of the line AB in the general form ax + by + c = 0. (6 marks)	AB = square root of (64 + 36) = 10 (2). Midpoint = (1, 1) (1). Gradient = -6/8 = -3/4 (1), hence 3x + 4y – 7 = 0 (2).

Section B: Circles, trigonometric equations and lines (Questions 5-8, 32 marks)

For equations, list every solution in the stated range and give a reason when a solution is rejected.

#	Question	Answer and marking notes
5	The circle C has equation x squared + y squared – 6x + 10y + 18 = 0. (a) Find its centre and radius. (b) Show that the point P(3, -1) lies on C. (c) Find the equation of the tangent to C at P. (8 marks)	(x – 3) squared + (y + 5) squared = 16, so the centre is (3, -5) and the radius is 4 (3). Substituting P gives 9 + 1 – 18 – 10 + 18 = 0, so P lies on C (2). CP is vertical because both points have x = 3, so the tangent is horizontal: y = -1 (3).
6	Solve 2 cos squared x + 3 sin x – 3 = 0 for x from 0 degrees to 360 degrees. (a) Reduce the equation to a quadratic equation in sin x. (b) Hence find all solutions in the given range. (8 marks)	Using cos squared x = 1 – sin squared x gives 2 sin squared x – 3 sin x + 1 = 0 (3), that is (2 sin x – 1)(sin x – 1) = 0 (2). sin x = 1/2 gives 30 and 150 degrees; sin x = 1 gives 90 degrees (2). So x = 30, 90 and 150 degrees (1).
7	The function is y = 3 sin 2x + 1. (a) State the amplitude, period, maximum value and minimum value. (b) Solve 3 sin 2x + 1 = 2.5 for x from 0 degrees to 180 degrees. (8 marks)	Amplitude 3 (1), period 180 degrees (1), maximum 4 (1), minimum -2 (1). sin 2x = 0.5 (1), so 2x = 30 or 150 degrees (2) and x = 15 or 75 degrees (1).
8	The line L has equation 4x – 3y + 5 = 0 and P is the point (2, 7). (a) Find the perpendicular distance from P to L. (b) Find the equation of the line through P perpendicular to L. (c) Find the coordinates of the foot of the perpendicular from P to L. (8 marks)	d = \|8 – 21 + 5\| / 5 = 1.6 (2). The gradient of L is 4/3 (1), so the perpendicular gradient is -3/4 (1) and the line is 3x + 4y – 34 = 0 (2). Solving the two equations gives the foot (3.28, 6.04) (2).

Section C: Three-dimensional trigonometry and applications (Questions 9-12, 44 marks)

In proofs give a reason at each step; in applied problems define the variables and state the assumptions made.

#	Question	Answer and marking notes
9	A cuboid measures 12 cm by 9 cm by 8 cm. (a) Find the length of a diagonal of the base. (b) Find the length of the space diagonal. (c) Find the angle between the space diagonal and the base. (d) Find the angle between the base diagonal and the 12 cm edge. (10 marks)	Base diagonal = 15 cm (2). Space diagonal = 17 cm (2). tan of the angle = 8/15, so the angle is 28.1 degrees (3). tan of the base angle = 9/12, giving 36.9 degrees (3).
10	Let f(x) = 3 sin x + 4 cos x. (a) Express f(x) in the form R sin(x + a), where R positive and a between 0 and 90 degrees. (b) Find the maximum value of f(x) and the value of x at which it occurs. (c) Solve f(x) = 2 for x from 0 degrees to 360 degrees. (11 marks)	R = 5 (1) and tan a = 4/3 gives a = 53.1 degrees (2), so f(x) = 5 sin(x + 53.1 degrees) (1). The maximum is 5 (1), occurring when x = 36.9 degrees (2). sin(x + 53.1 degrees) = 0.4 (1) gives x + 53.1 = 23.6, 156.4, 383.6 or 516.4 degrees (2), so x = 103.3 or 330.4 degrees in the range (1).
11	Given the points A(1, 2), B(7, 4) and C(5, 10). (a) Show that AB = BC and that angle ABC = 90 degrees. (b) Find the circumcentre of triangle ABC and the radius of its circumcircle. (c) Find the equation of the circumcircle. (11 marks)	AB squared = 40 and BC squared = 40, so AB = BC (2). AC squared = 80 = 40 + 40, so by the converse of Pythagoras angle ABC = 90 degrees (3). The circumcentre of a right-angled triangle is the midpoint of the hypotenuse: (3, 6) (2), with radius 4.47 (1). Circle: (x – 3) squared + (y – 6) squared = 20 (3).
12	From a point A on horizontal ground the angle of elevation of the top of a tower is 32 degrees. Walking 25 m from A straight towards the foot of the tower to a point B, the angle of elevation is 47 degrees. Let the horizontal distance from A to the foot of the tower be x metres. (a) Form an equation for the height of the tower in terms of x. (b) Find the height of the tower. (c) Find the distance from B to the top of the tower. (d) State one assumption used in the calculation. (12 marks)	h = x tan 32 and h = (x – 25) tan 47 (3), so x tan 32 = (x – 25) tan 47 (1). This gives x = 59.91 m (1) and h = 37.4 m (3). The distance from B to the foot is 34.91 m, so the distance from B to the top is the square root of (34.91 squared + 37.4 squared) = 51.2 m (3). Assumptions: the tower is vertical, the ground is horizontal and A, B and the foot are collinear (2).

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