Practice › Measures, Shape and Space › Basic Trigonometry

Basic Trigonometry: Paper 2 Multiple-Choice

Ten original Paper 2 questions on basic trigonometry: ratios in right-angled triangles, exact values, simplifying with sin²θ + cos²θ = 1 and the (90° − θ) identities, gradient, angles of depression, and lengths in a figure written in terms of an angle. Detailed solutions can be requested by email.

10 questions · about 15 minutes · answers below each question

Jump to

Question 1 — sin B from two legs of a right-angled triangle
Question 2 — exact value of tan 60° cos 30°
Question 3 — a base split by an altitude: AC in terms of h, α and β
Question 4 — simplifying (1 − sin²θ) tan θ
Question 5 — which statements about θ and 90° − θ must be true
Question 6 — two angles of depression from a building: distance between two cars
Question 7 — inclination of a road with gradient 1 : 5
Question 8 — a rectangle: AE ÷ DE using the equal opposite sides
Question 9 — two right-angled triangles sharing AC: AD in terms of ℓ and θ
Question 10 — value of (sin θ − cos θ) ÷ (sin θ + cos θ) when tan θ = 3

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 · Modelled on 2013 P2 Q22 — the same side-ratio idea, asking for one ratio instead of a ratio of two

△ A B C

∠ A C B = 9 0^{\circ}

A C = 5

and

B C = 12

. Find

sin B

$\frac{12}{13}$
$\frac{5}{12}$
$\frac{5}{13}$
$\frac{12}{5}$

Show answer

C. AB = 13 by Pythagoras' theorem. The side opposite B is AC = 5, so sin B = 5/13. Option A is sin A.

Question 2 · Modelled on 2013 P2 Q23 — exact values of special angles, in one product

tan 6 0^{\circ} cos 3 0^{\circ} =

$\frac{3}{2}$
$\frac{3}{2}$
$\frac{1}{2}$
$3$

Show answer

A. tan 60° = √3 and cos 30° = √3/2, so the product is 3/2.

Question 3 · Modelled on 2012 P2 Q18 — the same altitude figure, now asking for the base from the height

In the figure,

D

is a point lying on

A C

such that

B D ⊥ A C

. If

B D = h

∠ B A C = α

and

∠ B C A = β

, then

A C =

Triangle ABC with BD perpendicular to AC; angle at A is alpha, angle at C is beta, BD equals h

$h (tan α + tan β)$
$\frac{h}{tan α} + \frac{h}{tan β}$
$\frac{h}{sin α} + \frac{h}{sin β}$
$h tan α tan β$

Show answer

B. In △ABD, AD = h ÷ tan α; in △CBD, DC = h ÷ tan β. Option C adds the two slanting sides AB and BC instead.

Question 4 · New sub-type for Paper 2 — simplifying with sin²θ + cos²θ = 1 before anything else

(1 - sin^{2} θ) tan θ =

$cos θ$
$\frac{sin ^{2} θ}{cos θ}$
$sin θ$
$sin θ cos θ$

Show answer

D. 1 − sin²θ = cos²θ, and cos²θ × sin θ/cos θ = sin θ cos θ.

Question 5 · Modelled on 2013 P2 Q23 — 'which must be true' for acute angles, with a tempting false statement

0^{\circ} < θ < 9 0^{\circ}

, which of the following must be true?
I.

sin θ = cos (9 0^{\circ} - θ)

II.

tan θ tan (9 0^{\circ} - θ) = 1

III.

sin θ + cos θ = 1

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

Show answer

A. III confuses sin θ + cos θ with sin²θ + cos²θ. At θ = 45°, sin θ + cos θ = √2.

Question 6 · New sub-type for Paper 2 — two angles of depression, with distractors from the wrong ratio

From the top of a building

40 m

high, the angles of depression of two cars on level ground are

3 5^{\circ}

and

5 0^{\circ}

. The cars and the foot of the building lie in a straight line, with both cars on the same side of the building. Find the distance between the two cars, correct to the nearest

0.1 m

$17.5 m$
$19.7 m$
$23.6 m$
$33.6 m$

Show answer

C. The horizontal distances are 40 ÷ tan 35° and 40 ÷ tan 50°. Option B multiplies by tan instead of dividing; D is the nearer car only.

Question 7 · New sub-type for Paper 2 — gradient as a ratio, with options built from the wrong inverse function

The gradient of a straight road is

1 : 5

. The inclination of the road, correct to the nearest

0. 1^{\circ}

, is

$78. 7^{\circ}$
$11. 3^{\circ}$
$11. 5^{\circ}$
$78. 5^{\circ}$

Show answer

B. tan θ = 1/5 = 0.2. Option C uses sin θ = 0.2; options A and D give the angle with the vertical.

Question 8 · Modelled on 2019 P2 Q22 — a rectangle with two lines from one point; a ratio of slanting lengths

In the figure,

A B C D

is a rectangle and

E

is a point lying on

B C

. If

∠ A E B = α

and

∠ D E C = β

, then

\frac{A E}{D E} =

Rectangle ABCD with E on BC; lines EA and ED; angle AEB alpha, angle DEC beta

$\frac{sin α}{sin β}$
$\frac{sin β}{sin α}$
$\frac{cos α}{cos β}$
$\frac{cos β}{cos α}$

Show answer

B. AE = AB ÷ sin α and DE = DC ÷ sin β, and AB = DC. Dividing flips the sines.

Question 9 · Modelled on 2014 P2 Q18 — two right-angled triangles chained along AC, with equal angles at A

In the figure,

∠ A B C = ∠ A C D = 9 0^{\circ}

and

∠ B A C = ∠ C A D = θ

. If

A B = ℓ

, then

A D =

Two right-angled triangles ABC and ACD sharing side AC; right angles at B and C; both angles at A equal theta

$ℓ cos^{2} θ$
$ℓ tan^{2} θ$
$\frac{ℓ}{sin θ cos θ}$
$\frac{ℓ}{cos ^{2} θ}$

Show answer

D. AC = ℓ ÷ cos θ, then AD = AC ÷ cos θ. Each step divides by cos θ because the hypotenuse is the longer side.

Question 10 · New sub-type for Paper 2 — dividing through by cos θ to use a given tangent

tan θ = 3

, then

\frac{sin θ - cos θ}{sin θ + cos θ} =

$\frac{1}{2}$
$2$
$\frac{2}{3}$
$\frac{4}{5}$

Show answer

A. Divide top and bottom by cos θ: (tan θ − 1) ÷ (tan θ + 1) = 2 ÷ 4.

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

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.

Start with a free discovery session

No cost, no obligation. We look at where your child actually is, and show you what a term of real progress would look like.

Book a free discovery session See the programmes