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Basic Trigonometry: Paper 1 Long Questions

Ten Paper 1 questions on basic trigonometry: sine, cosine and tangent in right-angled triangles, exact values for 30°, 45° and 60°, sin²θ + cos²θ = 1 and the (90° − θ) identities, gradient, and angles of elevation and depression. Paper 1 rarely sets this topic as a question of its own, so most of the set has no direct past-paper model; it practises the steps that longer questions rely on. It opens with a single triangle and ends with every length in a figure written in terms of c and θ. Detailed solutions can be requested by email.

10 questions · 47 marks · about 55 minutes · answers below each question

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Question 1 — Pythagoras' theorem, then an angle from a tangent ratio
Question 2 — exact values of sin, cos and tan for 30°, 45° and 60°
Question 3 — simplifying with sin²θ + cos²θ = 1 and the (90° − θ) identities
Question 4 — from tan θ to exact sin θ and cos θ with a 5-12-13 triangle
Question 5 — two angles of elevation; height first, then the distance between two points
Question 6 — gradient 1 : 8 as an inclination; rise along a slope; a second road of different inclination
Question 7 — ladder against a wall: exact lengths at 60°, then a new angle after the foot slides
Question 8 — two right-angled triangles sharing a side; exact lengths and the area in surd form
Question 9 — two angles of depression from a cliff top; distance between two boats; a speed
Question 10 — the altitude to the hypotenuse; AD, CD and DB in terms of c and θ; proving CD² = AD × DB

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 sub-type, no direct past-paper model — a warm-up: Pythagoras' theorem, then an inverse ratio

△ A B C

∠ A C B = 9 0^{\circ}

A B = 25 cm

and

B C = 7 cm

(a)Find $A C$ . (1 mark)
(b)Find $∠ B A C$ , correct to the nearest $0. 1^{\circ}$ . (2 marks)

Show answer

(a) $24 cm$ (b) $16. 3^{\circ}$

Find AC first, then use tan A = 7 ÷ 24: both legs are exact, so no rounding error creeps in.

Question 2 · Modelled on 2013 P2 Q23 — exact values of the special angles, worked in full instead of judged as statements

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

(a) $sin 6 0^{\circ} cos 3 0^{\circ} + sin 3 0^{\circ} cos 6 0^{\circ}$ (1 mark)
(b) $(tan 4 5^{\circ} + tan 6 0^{\circ}) (tan 4 5^{\circ} - tan 6 0^{\circ})$ (1 mark)
(c) $2 sin^{2} 4 5^{\circ} - tan 3 0^{\circ} tan 6 0^{\circ}$ (1 mark)

Show answer

(a) $1$ (b) $- 2$ (c) $0$

(b) is a difference of two squares: 1² − (√3)². In (c), tan 30° and tan 60° are reciprocals.

Question 3 · Modelled on 2013 P2 Q23 — the (90° − θ) identities, here used to simplify rather than to test statements

Simplify each of the following, where

θ

is an acute angle.

(a) $\frac{1 - cos ^{2} θ}{sin θ cos θ}$ (2 marks)
(b) $tan θ sin (9 0^{\circ} - θ) + cos (9 0^{\circ} - θ)$ (2 marks)

Show answer

(a) $tan θ$ (b) $2 sin θ$

(a) 1 − cos²θ = sin²θ, then cancel one sin θ. (b) sin(90° − θ) = cos θ and cos(90° − θ) = sin θ.

Question 4 · Modelled on 2013 P2 Q22 — a ratio of sides turned into exact trigonometric ratios

It is given that

θ

is an acute angle and

tan θ = \frac{5}{12}

(a)Without using a calculator, find the values of $sin θ$ and $cos θ$ . (2 marks)
(b)Hence find the value of $\frac{sin θ + cos ( 9 0 ^{\circ} - θ )}{sin ( 9 0 ^{\circ} - θ )}$ . (2 marks)

Show answer

(a) $sin θ = \frac{5}{13}$ , $cos θ = \frac{12}{13}$ (b) $\frac{5}{6}$

Draw a right-angled triangle with sides 5 and 12; the hypotenuse is 13. In (b) the expression is 2 sin θ ÷ cos θ = 2 tan θ.

Question 5 · New sub-type, no direct past-paper model — two angles of elevation from the same side of a tower

In the figure,

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 and

B

lies between

A

and

F

A F = 100 m

. The angles of elevation of

T

from

A

and from

B

are

3 2^{\circ}

and

5 0^{\circ}

respectively.

Vertical tower TF with points A and B on the ground; angles of elevation 32 degrees at A and 50 degrees at B

(a)Find the height of the tower, correct to 3 significant figures. (2 marks)
(b)Find $A B$ , correct to 3 significant figures. (3 marks)

Show answer

(a) $62.5 m$ (b) $47.6 m$

Use △TFA first, where you know AF. Then FB = TF ÷ tan 50°, and AB = AF − FB. Keep the unrounded height for (b).

Question 6 · New sub-type, no direct past-paper model — gradient as a ratio, then lengths along and across a slope

A straight road has a gradient of

1 : 8

(a)Find the inclination of the road, correct to the nearest $0. 1^{\circ}$ . (1 mark)
(b)A car travels $1500 m$ along the road. Find the vertical rise of the car, correct to the nearest metre. (2 marks)
(c)Another straight road rises through the same vertical height, but its inclination is $6^{\circ}$ . Find the length of this second road, correct to 3 significant figures. (2 marks)

Show answer

(a) $7. 1^{\circ}$ (b) $186 m$ (c) $1780 m$

Gradient is vertical ÷ horizontal, so tan θ = 1/8. The 1500 m is measured along the road, so the rise is 1500 sin θ, not 1500 ÷ 8.

Question 7 · New sub-type, no direct past-paper model — a ladder that slides: exact values first, then a calculator

A ladder of length

6 m

leans against a vertical wall. The foot of the ladder is on level ground and the ladder makes an angle of

6 0^{\circ}

with the ground.

(a)Without using a calculator, find how high up the wall the ladder reaches, and how far the foot of the ladder is from the wall. (2 marks)
(b)The foot of the ladder now slides $1.2 m$ further away from the wall, and the top of the ladder stays on the wall. Find the new angle between the ladder and the ground, correct to the nearest $0. 1^{\circ}$ . (2 marks)
(c)How far does the top of the ladder slide down the wall? Give the answer correct to 3 significant figures. (2 marks)

Show answer

(a) $33 m$ up the wall; $3 m$ from the wall (b) $45. 6^{\circ}$ (c) $0.911 m$

After sliding, the foot is 3 + 1.2 = 4.2 m from the wall, so cos α = 4.2 ÷ 6. The ladder's length does not change.

Question 8 · Modelled on 2014 P2 Q18 — chained right-angled triangles, here with special angles and an area to finish

In the figure,

A B C D

is a quadrilateral with

∠ A B C = ∠ A C D = 9 0^{\circ}

A B = 12 cm

∠ B A C = 3 0^{\circ}

and

∠ C A D = 6 0^{\circ}

Quadrilateral ABCD split by diagonal AC; right angles at B and at C; angle BAC 30 degrees, angle CAD 60 degrees, AB 12 cm

(a)Without using a calculator, find $A C$ and $C D$ , giving the answers in surd form where necessary. (3 marks)
(b)Find the area of $A B C D$ in surd form. (3 marks)

Show answer

(a) $A C = 83 cm$ , $C D = 24 cm$ (b) $1203 cm^{2}$

AC is the hypotenuse of △ABC, so AC = 12 ÷ cos 30°. Then AC is the adjacent side in △ACD, so CD = AC tan 60°.

Question 9 · New sub-type, no direct past-paper model — angles of depression from a cliff, finishing with a speed

In the figure,

P Q

is a vertical cliff of height

120 m

, where

Q

is at sea level. Two boats

A

and

B

are on the sea, in a straight line with

Q

and on the same side of the cliff. The angles of depression of

A

and

B

from

P

are

4 0^{\circ}

and

2 5^{\circ}

respectively.

Vertical cliff PQ of height 120 m; boats A and B on the sea; angles of depression 40 degrees to A and 25 degrees to B

(a)Find $Q A$ and $QB$ , correct to the nearest metre. (2 marks)
(b)Find the distance between the two boats, correct to the nearest metre. (1 mark)
(c)Boat $B$ sails directly towards $A$ and reaches $A$ after $3$ minutes. Find the average speed of boat $B$ in metres per minute, correct to the nearest integer. (2 marks)

Show answer

(a) $Q A = 143 m$ , $QB = 257 m$ (b) $114 m$ (c) $38 m/min$

An angle of depression equals the angle of elevation from the boat (alternate angles), so QA = 120 ÷ tan 40°.

Question 10 · Modelled on 2021 P2 Q24 — the same chained right-angled triangles, every length written in terms of c and θ

In the figure,

∠ A C B = 9 0^{\circ}

and

D

is the point on

A B

such that

C D ⊥ A B

. Let

A B = c

and

∠ C A B = θ

Right-angled triangle ABC with right angle at C; CD perpendicular to AB; angle CAB marked theta

(a)Express $A C$ and $B C$ in terms of $c$ and $θ$ . (2 marks)
(b)Express $A D$ , $C D$ and $D B$ in terms of $c$ and $θ$ . (3 marks)
(c)Hence show that $C D^{2} = A D \times D B$ . (1 mark)

Show answer

(a) $A C = c cos θ$ , $B C = c sin θ$ (b) $A D = c cos^{2} θ$ , $C D = c sin θ cos θ$ , $D B = c sin^{2} θ$ (c) both sides equal $c^{2} sin^{2} θ cos^{2} θ$

∠BCD = θ too (both are 90° minus ∠ABC), so in △BCD, DB = BC sin θ. Adding AD + DB = c gives sin²θ + cos²θ = 1 for free.

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

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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