Right-angled trigonometry for HSC Maths Standard 2: sine, cosine and tangent, angles in degrees and minutes, elevation and depression (Year 12 Trigonometry, 2024 syllabus)

Syllabus dot point

“Apply the trigonometric ratios to right-angled triangles to find unknown side lengths or angles, in degrees and in degrees and minutes, and solve problems involving angles of elevation and depression”

9 min readUpdated 7 Oct 2026

Quick answer

Name the sides from the chosen angle, then use $\sin = \frac{\text{opp}}{\text{hyp}}$ , $\cos = \frac{\text{adj}}{\text{hyp}}$ or $\tan = \frac{\text{opp}}{\text{adj}}$ to find a side, or the inverse button to find an angle. Minutes are sixtieths of a degree. Angles of depression equal the alternate angle of elevation. Right-angled trigonometry opens Year 12 Trigonometry in the 2024 Mathematics Standard syllabus, before the sine rule, cosine rule and bearings.

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What this dot point is asking
The answer
Practice questions

What this dot point is asking

In the Mathematics Standard 11-12 Syllabus (2024), first examined in the 2027 HSC, Year 12 Standard 2 Trigonometry opens with right-angled triangles before it moves to the sine rule, cosine rule and area formula. NESA's content asks you to apply the trigonometric ratios to find unknown sides and angles "in both degrees, and degrees and minutes", and the later bearings and practical problems rely on them.

Note

Every right-angled triangle with the same angles has the same shape, only bigger or smaller. A ratio of two sides therefore depends only on the angle. Sine, cosine and tangent are just the names of the three ratios, and your calculator stores their values. Give it an angle and it returns the ratio. Give it the ratio and it returns the angle.

The answer

Key fact

Label the sides from the angle you are using: the hypotenuse is opposite the right angle, the opposite side faces the angle, the adjacent side touches it.

\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}, \qquad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}, \qquad \tan\theta = \frac{\text{opposite}}{\text{adjacent}}

Choose the ratio that contains the two sides you are dealing with (one known, one wanted, or both known when finding an angle).

Four steps every time

Draw the triangle and mark the angle and the sides.
Name the sides relative to that angle.
Pick the ratio that involves the known and wanted sides.
Solve. To find a side, multiply or divide. To find an angle, use $\sin^{-1}$ , $\cos^{-1}$ or $\tan^{-1}$ .

If the wanted side is the denominator, divide: $\tan 28^\circ = \frac{h}{40}$ gives $h = 40\tan 28^\circ$ , but $\tan 22^\circ = \frac{85}{d}$ gives $d = \frac{85}{\tan 22^\circ}$ .

Degrees and minutes

One degree is $60$ minutes, written $1^\circ = 60'$ . To convert $52^\circ 18'$ to degrees, compute $52 + \frac{18}{60} = 52.3^\circ$ . To convert an angle such as $61.9275^\circ$ , multiply the decimal part by $60$ : $0.9275 \times 60 = 55.65$ , so $61^\circ 56'$ to the nearest minute. Most scientific calculators have a degrees-minutes key that does this conversion directly.

Elevation and depression

The angle of elevation is measured up from the horizontal to the line of sight. The angle of depression is measured down from the horizontal. Because the two horizontals are parallel, the angle of depression from the top equals the angle of elevation from the bottom (alternate angles), so a depression problem becomes an ordinary right-angled triangle once you move the angle inside it.

Worked examples

A side from an angle

A ramp $12$ m long rises at $35^\circ$ . The height is opposite the angle, so $12\sin 35^\circ \approx 6.9$ m, and the horizontal distance is adjacent, so $12\cos 35^\circ \approx 9.8$ m.

Marker's note: name the sides first, then the ratio, then the answer with units.

An angle from two sides

Legs of $15$ cm (opposite) and $8$ cm (adjacent): $\tan\theta = \frac{15}{8}$ , $\theta \approx 61.9^\circ$ , which is $61^\circ 56'$ to the nearest minute.

Marker's note: do not round $\frac{15}{8}$ before using $\tan^{-1}$ .

Elevation with eye height

An observer $40$ m away sees the top of a pole at $28^\circ$ elevation, with eyes $1.6$ m high. The pole above eye level is $40\tan 28^\circ = 21.27$ m, so the pole is about $22.9$ m.

Marker's note: add the eye height, because the triangle starts at the observer's eyes, not the ground.

Common traps

Calculator in radian mode: $\sin 35$ should give $0.5736$ , not $-0.4282$ .
Labelling sides from the wrong angle: Opposite and adjacent swap when you change angle. Only the hypotenuse stays fixed.
Multiplying when you should divide: If the unknown is on the bottom of the ratio, rearrange before you calculate.
Treating $52^\circ 18'$ as $52.18^\circ$: Minutes are out of $60$ , not $100$ .
Rounding too early: Carry full calculator values and round only the final answer.

Exam technique

Redraw the diagram larger than the one printed and put the angle at the observer. For depression questions, copy the angle to the bottom position as an elevation. State the units and the rounding the question asks for ("to the nearest degree", "to one decimal place", "to the nearest minute").

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

A ramp is

12

m long and rises at

35^\circ

to the horizontal. Find, correct to one decimal place, the height the ramp rises and the horizontal distance it covers.

Show worked solution →

Choose the ratios. The $12$ m ramp is the hypotenuse. The height is opposite the $35^\circ$ angle and the horizontal distance is adjacent to it.

\text{height} = 12 \sin 35^\circ = 6.88\ldots \approx 6.9 \text{ m}

\text{horizontal distance} = 12 \cos 35^\circ = 9.83\ldots \approx 9.8 \text{ m}

Marker's note: one mark for each correct ratio and answer. Check with Pythagoras: $6.88^2 + 9.83^2 = 144.0$ , so the hypotenuse is $12$ .

foundation2 marks

A right-angled triangle has an opposite side of

15

cm and an adjacent side of

8

cm to the angle

\theta

. Find

\theta

to the nearest degree, and then to the nearest minute.

Show worked solution →

Choose the ratio. Opposite and adjacent are known, so use tangent.

\tan\theta = \frac{15}{8}, \qquad \theta = \tan^{-1}\left(\frac{15}{8}\right) = 61.927\ldots^\circ

Nearest degree: $\theta \approx 62^\circ$ .

Nearest minute. The decimal part is $0.9275 \times 60 = 55.65$ , which rounds to $56$ minutes, so $\theta \approx 61^\circ 56'$ .

Marker's note: one mark for $\tan^{-1}\left(\frac{15}{8}\right)$ and one for both rounded answers. The hypotenuse is $\sqrt{15^2 + 8^2} = 17$ cm.

core3 marks

From a point on level ground

40

m from the base of a flagpole, the angle of elevation of the top of the pole is

28^\circ

. The observer's eyes are

1.6

m above the ground. Find the height of the flagpole to one decimal place.

Show worked solution →

Draw the triangle at eye level. The horizontal side is $40$ m and the angle at the eye is $28^\circ$ . The height above eye level is opposite the angle.

\tan 28^\circ = \frac{h}{40}, \qquad h = 40 \tan 28^\circ = 21.27\ldots \text{ m}

Add the eye height. $21.27 + 1.6 = 22.87$ , so the flagpole is about $22.9$ m tall.

Marker's note: one mark for the tangent equation, one for $21.27$ , one for adding $1.6$ . Forgetting the eye height is the usual lost mark.

core3 marks

6.5

m ladder reaches

5.8

m up a vertical wall. Find the angle the ladder makes with the ground to the nearest degree, and how far the foot of the ladder is from the wall to one decimal place.

Show worked solution →

Angle. The wall height is opposite the angle at the ground and the ladder is the hypotenuse, so use sine.

\sin\theta = \frac{5.8}{6.5}, \qquad \theta = \sin^{-1}\left(\frac{5.8}{6.5}\right) = 63.16\ldots^\circ \approx 63^\circ

Distance from the wall. By Pythagoras,

d = \sqrt{6.5^2 - 5.8^2} = \sqrt{8.61} = 2.93\ldots \approx 2.9 \text{ m}.

Marker's note: one mark for the sine ratio, one for $63^\circ$ , one for $2.9$ m. Using the unrounded angle gives the same distance: $6.5\cos 63.16^\circ = 2.93$ m.

exam4 marks

From the top of a vertical cliff

85

m above sea level, the angle of depression of a boat is

22^\circ

. A few minutes later the angle of depression of the boat is

41^\circ

, and the boat has moved straight towards the base of the cliff. How far did the boat travel, to the nearest metre?

Show worked solution →

Use the angle inside the triangle: The angle of depression is measured from the horizontal, so by alternate angles it equals the angle of elevation of the cliff top from the boat. In each position the cliff height $85$ m is opposite that angle and the horizontal distance is adjacent.
First position: $d_1 = \frac{85}{\tan 22^\circ} = 210.38\ldots$ m.
Second position: $d_2 = \frac{85}{\tan 41^\circ} = 97.78\ldots$ m.
Distance travelled: $210.38 - 97.78 = 112.60\ldots \approx 113$ m.

Marker's note: one mark for recognising the alternate angle, one mark for each distance, one for the subtraction. Round only at the end, since rounding $d_1$ and $d_2$ first can shift the answer by a metre.

exam4 marks

(a) A right-angled triangle has legs of

5.2

m and

3.7

m. Find the smaller acute angle to the nearest minute. (b) In another right-angled triangle the hypotenuse is

9.4

m and one angle is

52^\circ 18'

. Find the side opposite that angle to one decimal place.

Show worked solution →

(a) Smaller angle. The smaller angle is opposite the shorter leg.

\tan\theta = \frac{3.7}{5.2}, \qquad \theta = 35.433\ldots^\circ

The decimal part gives $0.4333 \times 60 = 26.0$ minutes, so $\theta \approx 35^\circ 26'$ .

(b) Opposite side. Opposite and hypotenuse give sine. Convert the angle first (or enter it with the degrees-minutes key): $52^\circ 18' = 52 + \frac{18}{60} = 52.3^\circ$ .

\text{opposite} = 9.4 \sin 52.3^\circ = 7.44\ldots \approx 7.4 \text{ m}

Marker's note: two marks for (a), two for (b). Check that the calculator is in degree mode.

Practise this

Sources & how we know this

Mathematics Standard 11-12 Syllabus (2024): Year 12 Standard 2 Trigonometry — NESA (2024)
Mathematics Standard Stage 6 Syllabus (2017) — NESA
Mathematics Standard HSC exam papers — NESA