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Basic Trigonometry: Paper 1 Challenge

Ten harder Paper 1 questions on basic trigonometry. Each one states a condition rather than naming a method: a ratio hidden in an equation, long sums collapsed by pairing θ with 90° − θ, a triangle fixed by its perimeter, a drone whose angle of elevation does not rise steadily, and a ramp rule that depends on the horizontal run. Work the standard Paper 1 set first. Detailed solutions can be requested by email.

10 questions · 48 marks · about 60 minutes · answers below each question

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Question 1 — sin θ = 2 cos θ: exact values with no triangle given
Question 2 — proving an identity with reciprocals; using it at 30°
Question 3 — pairing θ with 90° − θ: a sum of 89 squares and a product of 89 tangents
Question 4 — perimeter plus one ratio fixes a right-angled triangle; testing a claim about two areas
Question 5 — D the mid-point of BC: show tan ∠BAC = 2 tan α, then find β
Question 6 — a 15% road sign as an inclination; rise along the road; testing a map-distance claim
Question 7 — two angles of elevation and AB only: an equation for the height in surd form
Question 8 — a rising drone: speed from two angles of elevation; when the angle reaches 45°; testing a 'steady angle' claim
Question 9 — proving 1 < sin θ + cos θ ≤ √2 for every acute angle
Question 10 — a ramp rule of 1 : 12; least run and least length; a trap that uses the sloping length

All questions are original, written for Math and AI Academy and modelled on past-paper style. They are not reproductions of HKEAA examination questions.

Question 1 · New technique — the ratio is hidden in an equation; no triangle and no value of tan θ is given

It is given that

θ

is an acute angle and

sin θ = 2 cos θ

(a)Without using a calculator, find the exact values of $sin θ$ and $cos θ$ , with rational denominators. (2 marks)
(b)Hence find the values of $sin θ cos θ$ and $(sin θ + cos θ)^{2}$ . (2 marks)

Show answer

(a) $sin θ = \frac{2 5}{5}$ , $cos θ = \frac{5}{5}$ (b) $\frac{2}{5}$ and $\frac{9}{5}$

Divide by cos θ to get tan θ = 2, or substitute sin θ = 2 cos θ into sin²θ + cos²θ = 1. Both cos θ and sin θ are positive because θ is acute.

Question 2 · New technique — a two-sided identity to prove, then used to evaluate without a calculator

Let

θ

be an acute angle.

(a)Prove that $(\frac{1}{cos θ} - cos θ) (\frac{1}{sin θ} - sin θ) = sin θ cos θ$ . (3 marks)
(b)Hence, without using a calculator, find the value of $(\frac{1}{cos 3 0 ^{\circ}} - cos 3 0^{\circ}) (\frac{1}{sin 3 0 ^{\circ}} - sin 3 0^{\circ})$ . (1 mark)

Show answer

(a) Each bracket simplifies: $\frac{sin ^{2} θ}{cos θ} \cdot \frac{cos ^{2} θ}{sin θ} = sin θ cos θ$ (b) $\frac{3}{4}$

Combine each bracket into one fraction: 1/cos θ − cos θ = (1 − cos²θ)/cos θ = sin²θ/cos θ.

Question 3 · New technique — pairing complementary angles to collapse a long sum and a long product

Without using a calculator, find the value of each of the following.

(a) $sin^{2} 1^{\circ} + sin^{2} 2^{\circ} + sin^{2} 3^{\circ} + \dots + sin^{2} 8 9^{\circ}$ (2 marks)
(b) $tan 1^{\circ} \times tan 2^{\circ} \times tan 3^{\circ} \times \dots \times tan 8 9^{\circ}$ (2 marks)

Show answer

(a) $\frac{89}{2}$ (b) $1$

sin²(90° − θ) = cos²θ, so sin²1° + sin²89° = 1, and so on for 44 pairs, with sin²45° = 1/2 left over. For (b), tan θ × tan(90° − θ) = 1.

Question 4 · New technique — a trigonometric ratio fixes the shape, the perimeter fixes the size; then a claim to test

A right-angled triangle has a perimeter of

60 cm

(a)If the tangent of one of its acute angles is $\frac{5}{12}$ , find the area of the triangle. (2 marks)
(b)Another right-angled triangle also has a perimeter of $60 cm$ , and the sine of one of its acute angles is $\frac{3}{5}$ . Someone claims that the two triangles have the same area. Do you agree? Explain your answer. (3 marks)

Show answer

(a) $120 cm^{2}$ (b) Disagree: the second triangle has sides 15, 20, 25 cm and area $150 cm^{2}$

tan = 5/12 means the sides are 5k, 12k and 13k, so 30k = 60. For sine 3/5 the sides are 3k, 4k and 5k, so 12k = 60.

Question 5 · New technique — a mid-point condition replaces a given length; the student must find a second angle

In the figure,

∠ A B C = 9 0^{\circ}

and

D

is the mid-point of

B C

. Let

∠ B A D = α

and

∠ D A C = β

Right-angled triangle ABC with right angle at B; D the mid-point of BC; angle BAD alpha and angle DAC beta

(a)Show that $tan ∠ B A C = 2 tan α$ . (2 marks)
(b)If $α = 2 0^{\circ}$ , find $β$ , correct to the nearest $0. 1^{\circ}$ . (3 marks)

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(a) $tan ∠ B A C = \frac{B C}{A B} = \frac{2 B D}{A B} = 2 tan α$ (b) $16. 1^{\circ}$

Both angles share the side AB. Since BC = 2BD, tan ∠BAC = 2 tan α. Then β = ∠BAC − α, and β is not equal to α, even though BD = DC.

Question 6 · New technique — a road-sign percentage converted to a gradient; a claim that mixes up slope and map distance

A road sign says that a straight road has a gradient of

15%

. A cyclist rides

2 km

along the road, from the bottom to the top.

(a)Find the inclination of the road, correct to the nearest $0. 1^{\circ}$ . (1 mark)
(b)Find the vertical rise of the cyclist, correct to the nearest metre. (2 marks)
(c)A map shows horizontal distances only. Someone claims that this $2 km$ stretch of road also measures $2 km$ on the map. Do you agree? Explain your answer, giving the map distance correct to the nearest metre. (2 marks)

Show answer

(a) $8. 5^{\circ}$ (b) $297 m$ (c) Disagree: the horizontal distance is about $1978 m$

15% means rise ÷ run = 0.15, so tan θ = 0.15. The 2 km is along the slope: rise = 2000 sin θ, map distance = 2000 cos θ.

Question 7 · New technique — the height is unknown in both triangles; set up one equation and solve for it

T F

is a vertical tower standing on level ground.

A

and

B

are two points on the ground such that

A

B

and

F

are collinear,

B

lies between

A

and

F

, and

A B = 50 m

. The angles of elevation of

T

from

A

and from

B

are

3 0^{\circ}

and

6 0^{\circ}

respectively.

(a)Without using a calculator, find the height of the tower in surd form. (3 marks)
(b)Show that $B T = A B$ . (2 marks)

Show answer

(a) $253 m$ (b) $B T = \frac{25 3}{sin 6 0 ^{\circ}} = 50 m = A B$

Let the height be h. Then AF = h ÷ tan 30° = √3 h and BF = h ÷ tan 60° = h/√3, and AF − BF = 50.

Question 8 · New technique — angles of elevation as a rate; a claim that treats the angle as rising steadily

A drone rises vertically at a constant speed above a point

F

on level ground. An observer stands at a point

O

on the ground,

60 m

from

F

. At a certain moment the angle of elevation of the drone from

O

2 0^{\circ}

. Exactly

10

seconds later it is

3 5^{\circ}

(a)Find the speed of the drone in m/s, correct to 3 significant figures. (3 marks)
(b)Find how many seconds after the first moment the angle of elevation reaches $4 5^{\circ}$ , correct to 3 significant figures. (2 marks)
(c)A student argues: 'The angle rose by $1 5^{\circ}$ in $10$ seconds, so it rises by another $1 0^{\circ}$ in about $6.7$ seconds.' Explain why the student's method is wrong. (1 mark)

Show answer

(a) $2.02 m/s$ (b) $18.9 s$ (c) The height rises steadily but the angle does not; the true time after the second moment is about $8.9 s$ , not $6.7 s$

The heights are 60 tan 20° and 60 tan 35°. At 45° the height is exactly 60 m. The angle grows more and more slowly as the drone climbs.

Question 9 · New technique — bounds on sin θ + cos θ proved from the identity, with no calculator

Let

θ

be an acute angle.

(a)Show that $(sin θ + cos θ)^{2} + (sin θ - cos θ)^{2} = 2$ . (2 marks)
(b)Hence show that $sin θ + cos θ \leq 2$ . (1 mark)
(c)Show also that $sin θ + cos θ > 1$ . (1 mark)

Show answer

(a) Expand: the cross terms cancel, leaving $2 (sin^{2} θ + cos^{2} θ) = 2$ (b) $(sin θ + cos θ)^{2} \leq 2$ because the other square is never negative (c) $(sin θ + cos θ)^{2} = 1 + 2 sin θ cos θ > 1$

For (c), sin θ and cos θ are both positive for an acute angle, so 2 sin θ cos θ is positive.

Question 10 · New technique — a building rule that depends on the horizontal run, with a near-miss trap

A wheelchair ramp must rise

0.6 m

. A building rule says that the gradient of the ramp must not be steeper than

1 : 12

(a)Find the least possible horizontal distance covered by the ramp. (1 mark)
(b)Find the least possible length of the ramp's surface, correct to 3 significant figures, and the inclination of that ramp, correct to the nearest $0. 1^{\circ}$ . (2 marks)
(c)A contractor builds a ramp whose surface is $7.21 m$ long and rises $0.6 m$ . He says: 'Since $0.6 \div 7.21$ is less than $\frac{1}{12}$ , the ramp obeys the rule.' Do you agree? Explain your answer. (3 marks)

Show answer

(a) $7.2 m$ (b) $7.22 m$ ; $4. 8^{\circ}$ (c) Disagree: the horizontal run is about $7.185 m$ , so the gradient is about $0.0835$ , steeper than $\frac{1}{12} \approx 0.0833$

Gradient is rise ÷ horizontal run, not rise ÷ sloping length. 0.6 ÷ 7.21 is the sine of the inclination, not the tangent.

Continue practising this topic: Try the Paper 2 Challenge MC for harder variants, or move to Paper 1 long questions — also available as Paper 2 multiple-choice.

Where this sits in the syllabus

Basic trigonometry is a junior topic in the Measures, Shape and Space strand. In HKDSE Paper 2 it has appeared as one multiple-choice question in most years from 2012 to 2025, usually a figure built from right-angled triangles in which one length must be written in terms of another length and an angle. On Paper 1 it rarely stands alone; it turns up as a step inside longer questions. Most marks are lost in two places: using the hypotenuse where the adjacent side is needed, and rounding an intermediate length too early.

Full coverage of this topic, and the rest of the course, is on the Mathematics page. More sets are listed on the practice index.

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