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Basic Trigonometry: Paper 2 Multiple-Choice Challenge

Ten harder Paper 2 questions on basic trigonometry. Each wrong option is a real mistake: forgetting a square root, treating a distance along a slope as horizontal, assuming that equal steps in a tangent mean equal steps in the angle, or putting both observers on the same side of a tower. Work the standard Paper 2 set first. Detailed solutions can be requested by email.

10 questions · about 20 minutes · answers below each question

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Question 1 — sin θ + cos θ from sin θ cos θ
Question 2 — cos²10° + cos²20° + … + cos²80° by pairing
Question 3 — BD = DE = EC: which statements about the angles at A are true
Question 4 — tan 20° tan 40° tan 50° tan 70° without a calculator
Question 5 — two angles of elevation, 30° and 45°, and AB = 20 m: the height in surd form
Question 6 — rise of a car travelling 500 m along a road of gradient 1 : 10
Question 7 — a square with AE ⊥ EF: CF in terms of a and θ
Question 8 — 1 ÷ tan θ in terms of k when sin θ = k
Question 9 — which inequalities hold for every acute angle
Question 10 — tower between two observers: height from 40°, 25° and AB = 100 m

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 — sin θ cos θ given; the square of a sum, with the difference as a distractor

It is given that

θ

is an acute angle and

sin θ cos θ = \frac{12}{25}

. Find

sin θ + cos θ

$\frac{1}{5}$
$\frac{7}{5}$
$\frac{24}{25}$
$\frac{49}{25}$

Show answer

B. (sin θ + cos θ)² = 1 + 2 × 12/25 = 49/25. Option D forgets the square root; option A is the size of sin θ − cos θ.

Question 2 · New technique — complementary pairs collapse a sum of eight squares

cos^{2} 1 0^{\circ} + cos^{2} 2 0^{\circ} + cos^{2} 3 0^{\circ} + \dots + cos^{2} 8 0^{\circ} =

$\frac{7}{2}$
$4$
$\frac{9}{2}$
$8$

Show answer

B. Pair 10° with 80°, 20° with 70°, and so on: each pair is cos²θ + sin²θ = 1. There are 4 pairs and no middle term.

Question 3 · Modelled on 2018 P2 Q21 — points dividing a side into equal parts; equal tangents do not mean equal angles

In the figure,

∠ A B C = 9 0^{\circ}

D

and

E

are points lying on

B C

such that

B D = D E = E C

. Which of the following must be true?
I.

tan ∠ B A E = 2 tan ∠ B A D

II.

∠ B A E = 2∠ B A D

III.

tan ∠ B A C = tan ∠ B A D + tan ∠ B A E

Right-angled triangle ABC with right angle at B; D and E divide BC into three equal parts; lines AD and AE

I only
I and II only
I and III only
I, II and III

Show answer

C. With BD = u: tan ∠BAD = u/AB, tan ∠BAE = 2u/AB, tan ∠BAC = 3u/AB. So I and III hold. Tangents double, but angles do not.

Question 4 · New technique — a product of four tangents collapsed by complementary pairs

tan 2 0^{\circ} tan 4 0^{\circ} tan 5 0^{\circ} tan 7 0^{\circ} =

$1$
$\frac{3}{3}$
$3$
$3$

Show answer

A. tan 70° = 1/tan 20° and tan 50° = 1/tan 40°, so every factor cancels.

Question 5 · New technique — the height is the unknown in an equation; a surd answer with a rationalized denominator

A

and

B

are two points on level ground,

20 m

apart, in a straight line with the foot of a vertical pole. The angles of elevation of the top of the pole from

A

and

B

are

3 0^{\circ}

and

4 5^{\circ}

respectively. Find the height of the pole.

$10 (3 - 1) m$
$103 m$
$20 (3 + 1) m$
$10 (3 + 1) m$

Show answer

D. Let the height be h. The distances to the foot are √3 h and h, so √3 h − h = 20 and h = 20 ÷ (√3 − 1).

Question 6 · New technique — distance along a slope versus horizontal distance; the obvious answer is wrong

A straight road has a gradient of

1 : 10

. A car travels

500 m

along the road. Find the vertical rise of the car, correct to the nearest

0.1 m

$49.8 m$
$50.0 m$
$50.3 m$
$497.5 m$

Show answer

A. The 500 m is measured along the road. tan θ = 0.1, so the rise is 500 sin θ ≈ 49.8 m. Option B treats 500 m as the horizontal run; D is the horizontal run.

Question 7 · Modelled on 2023 P2 Q23 — a right angle placed inside a square; the student must find the equal angle

In the figure,

A B C D

is a square of side

a

E

and

F

are points lying on

B C

and

C D

respectively such that

∠ A E F = 9 0^{\circ}

. If

∠ B A E = θ

, then

C F =

Square ABCD with E on BC and F on CD; angle AEF is a right angle; angle BAE is theta

$a (1 - tan θ)$
$a tan^{2} θ$
$a tan θ (1 - tan θ)$
$\frac{a ( 1 - tan θ )}{tan θ}$

Show answer

C. BE = a tan θ, so EC = a − a tan θ. ∠CEF = θ as well (angles on a straight line at E), so CF = EC tan θ.

Question 8 · New technique — a ratio given as a letter; the student builds the triangle in terms of k

θ

is an acute angle and

sin θ = k

, then

\frac{1}{tan θ} =

$\frac{k}{1 - k ^{2}}$
$\frac{1 - k ^{2}}{k}$
$\frac{1}{1 - k ^{2}}$
$1 - k^{2}$

Show answer

B. cos θ = √(1 − k²), so 1/tan θ = cos θ ÷ sin θ = √(1 − k²) ÷ k. Option A is tan θ itself.

Question 9 · Modelled on 2013 P2 Q23 — 'must be true' statements where one fails only for some angles

0^{\circ} < θ < 9 0^{\circ}

, which of the following must be true?
I.

sin θ < tan θ

II.

sin θ + cos θ > 1

III.

sin θ tan θ < 1

I only
II only
I and III only
I and II only

Show answer

D. I: tan θ = sin θ ÷ cos θ and 0 < cos θ < 1. II: the square is 1 + 2 sin θ cos θ. III fails at 60°: (√3/2)(√3) = 3/2.

Question 10 · New technique — the observers stand on opposite sides of the tower; three tempting wrong models as options

A

and

B

are two points on level ground,

100 m

apart. A vertical tower stands at a point on

A B

between

A

and

B

. The angles of elevation of the top of the tower from

A

and

B

are

4 0^{\circ}

and

2 5^{\circ}

respectively. Find the height of the tower, correct to the nearest

0.1 m

$30.0 m$
$31.9 m$
$46.6 m$
$105 m$

Show answer

A. The two horizontal distances add up to 100 m: h ÷ tan 40° + h ÷ tan 25° = 100. Option D subtracts them, as if A and B were on the same side.

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

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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