S4 Mathematics Paper: Trigonometric Ratios And Applications (100 Marks)

S4 Mathematics Paper: Trigonometric Ratios and Applications (100 marks)

2026-11-11

S4 Mathematics – trigonometric ratios and applications

S4 mathematics practice paper on trigonometric ratios, special angles, angles of elevation, the cosine rule, triangle area and bearings, with answers.

Download　Question paper PDF　|　Answer key PDF

Paper details　S4 mathematics practice paper on trigonometric ratios and applications: special angles, elevation, cosine rule, triangle area, bearings and trigonometric equations, with answers.

Part A Question 1: Trigonometric Ratios and Special Angles

Simplify all answers and state the trigonometric ratio used where necessary.

#	Question	Answer and marking notes
1	(a) In a right-angled triangle, the side opposite to angle theta is 8 cm and the hypotenuse is 17 cm. Find the values of sin theta and cos theta.	Answer: sin theta = 8/17, cos theta = 15/17. By Pythagoras, the adjacent side is the square root of (17 squared minus 8 squared), which gives 15 cm, so cos theta = 15/17. Marking: adjacent side 1, sin 1, cos 1, fraction form 1.
2	(b) Without using a calculator, evaluate sin 30 degrees + cos 60 degrees + tan 45 degrees.	Answer: 2. sin 30 degrees = 1/2, cos 60 degrees = 1/2, tan 45 degrees = 1, so the total is 1/2 + 1/2 + 1 = 2. Marking: each special angle 1, total 1.
3	(c) Given tan theta = 3/4 and theta is an acute angle, find sin theta and cos theta.	Answer: sin theta = 3/5, cos theta = 4/5. Since tan theta is opposite over adjacent, take opposite 3 and adjacent 4, so the hypotenuse is 5, giving sin theta = 3/5 and cos theta = 4/5. Marking: hypotenuse 1, each ratio 1, note that acute angle takes positive values 1.
4	(d) Evaluate sin squared 35 degrees plus cos squared 35 degrees plus tan 45 degrees.	Answer: 2. For the same acute angle the identity sin squared theta plus cos squared theta equals 1, and tan 45 degrees equals 1, so the total is 2. Marking: identity 2, tan 45 degrees 1, total 1.

Part A Question 2: Angle of Elevation in a Real Measurement

From a point A on horizontal ground, the angle of elevation of the top of a vertical tower is 28 degrees, and the horizontal distance from A to the foot of the tower is 60 m.

#	Question	Answer and marking notes
5	(a) Find the height of the tower, correct to one decimal place.	Answer: about 31.9 m. tan 28 degrees equals height over 60, so the height is 60 tan 28 degrees, which is 31.9 m. Marking: correct ratio 1, substitution 1, value 1, one decimal place 1.
6	(b) Find the distance along the line of sight from A to the top of the tower, correct to one decimal place.	Answer: about 68.0 m. cos 28 degrees equals 60 over the line of sight, so the distance is 60 divided by cos 28 degrees, which is 68.0 m. Marking: cosine relation 2, value 1, unit 1.
7	(c) If the observation point moves to B, 40 m straight in front of the tower, find the new angle of elevation, correct to one decimal place.	Answer: about 38.6 degrees. tan theta equals 31.9 over 40, so theta is 38.6 degrees. Marking: formula 2, value 1, one decimal place 1.
8	(d) Compare the angles of elevation measured at A and at B, and explain the difference.	Answer: the angle at B is larger. The height does not change while the horizontal distance falls from 60 m to 40 m, and tan theta is height divided by horizontal distance, so a smaller denominator gives a larger ratio and a larger angle. Marking: B is larger 2, link to tan theta 2.

Part A Question 3: Cosine Rule and Area of a Triangle

In triangle ABC, BC = 7 cm, AC = 9 cm and angle C = 42 degrees.

#	Question	Answer and marking notes
9	(a) Find the length of AB, correct to two decimal places.	Answer: about 6.03 cm. By the cosine rule, AB squared equals 49 + 81 minus 2 times 7 times 9 times cos 42 degrees, which is 130 minus 93.64, giving 36.36, so AB is 6.03 cm. Marking: correct formula 2, substitution 1, value 1.
10	(b) Find the area of triangle ABC, correct to one decimal place.	Answer: about 21.1 square centimetres. Area equals half times 7 times 9 times sin 42 degrees, which gives 21.1. Marking: formula 2, value 1, unit 1.
11	(c) Find angle A, correct to one decimal place.	Answer: about 51.0 degrees. By the sine rule, sin A over 7 equals sin 42 degrees over 6.03, so sin A is 0.7767 and A is 51.0 degrees. Marking: sine rule 2, substitution 1, value 1.
12	(d) Hence find angle B and check the angle sum of the triangle.	Answer: B is about 87.0 degrees. 180 minus 42 minus 51.0 gives 87.0, and the three angles add to 180, so the check holds. Marking: value of B 2, angle sum check 2.

Part A Question 4: Bearings and Displacement

A ship leaves port P and sails 12 km on a bearing of 060 degrees, then 9 km on a bearing of 150 degrees.

#	Question	Answer and marking notes
13	(a) Find the straight-line distance between the ship and port P, correct to one decimal place.	Answer: about 15.0 km. Resolve both legs into east and north components: (10.39, 6.00) plus (4.50, minus 7.79) gives (14.89, minus 1.79), so the distance is the square root of (14.89 squared plus 1.79 squared), which is 15.0 km. Marking: components 2, squares 1, value 1.
14	(b) Find the bearing of the ship from port P, to the nearest degree.	Answer: about 097 degrees. tan theta equals 14.89 over 1.79, and with a positive east component and a negative north component the bearing is 097 degrees. Marking: tangent relation 1, quadrant reasoning 2, degrees 1.
15	(c) The ship actually travelled 21 km but is only about 15 km from the port. Explain why the two values differ.	Answer: 21 km is the distance travelled along the path while 15 km is the magnitude of the displacement. The two legs point in different directions, so the second partly cancels the first and the straight-line distance is shorter than the total travelled. Marking: path versus displacement 2, different directions 1, cancellation 1.
16	(d) If the second leg were 9 km on a bearing of 240 degrees instead, find the distance from P, correct to one decimal place.	Answer: about 3.0 km. The new components are (minus 7.79, minus 4.50), giving a resultant of (2.60, 1.50), so the distance is the square root of (2.60 squared plus 1.50 squared), which is 3.0 km. Marking: components 2, resultant 1, value 1.

Part B Question 5: Trigonometric Equations

All angles in this question are in degrees and solutions lie between 0 and 360 degrees inclusive.

#	Question	Answer and marking notes
17	(a) Solve 2 sin squared x minus sin x minus 1 equals 0 by first finding the possible values of sin x.	Answer: sin x = 1 or sin x = minus 1/2. Factorise the equation as (2 sin x + 1)(sin x minus 1) equals 0, so sin x is minus 1/2 or 1. Marking: factorisation 3, each value 1, treating sin x as the unknown 1.
18	(b) Hence find all solutions of the equation in the given range.	Answer: x equals 90, 210 and 330 degrees. sin x = 1 gives 90 degrees, while sin x = minus 1/2 gives 210 and 330 degrees in the third and fourth quadrants. Marking: 90 degrees 1, 210 degrees 2, 330 degrees 2, complete ordered list 1.
19	(c) Solve cos 2x = 0.5 for x between 0 and 360 degrees.	Answer: x equals 30, 150, 210 and 330 degrees. Since 2x ranges from 0 to 720 degrees, it takes the values 60, 300, 420 and 660 degrees, and each is halved. Marking: four values of 2x 3, halving 3.

Part B Question 6: Trigonometric Identity and Its Use

Full working is required in this question; answers alone do not earn full marks.

#	Question	Answer and marking notes
20	(a) Prove that (1 minus cos squared theta) divided by sin theta equals sin theta for any angle theta with sin theta not equal to zero.	Answer: (1 minus cos squared theta) over sin theta equals sin squared theta over sin theta, which is sin theta, using the identity sin squared theta plus cos squared theta equals 1. Marking: identity 2, factorising 2, cancelling with the non-zero condition stated 2.
21	(b) Hence solve (1 minus cos squared x) divided by sin x equals 0.5 for x between 0 and 360 degrees.	Answer: x equals 30 and 150 degrees. From part (a), sin x equals 0.5, which gives 30 and 150 degrees in the first and second quadrants. Marking: reducing to sin x = 0.5 gets 2, each solution 2.
22	(c) Explain why x = 0, 180 and 360 degrees cannot be solutions of part (b).	Answer: at these three angles the sine is zero, so the denominator sin x is zero and the expression is undefined, which means they cannot be substituted; 0.5 cannot equal an undefined quantity. Marking: sine equals zero 2, undefined denominator 2, clear conclusion 2.

Find a tutor — free Tutor Registration →

A TutorZone practice paper; original questions and passages, not publisher material.