Skip to main content
ExamExplained

Right-angled triangles, trigonometry, elevation, depression and bearings: HSC Maths Standard 1 Year 12

Syllabus dot point

“Solve practical problems involving right-angled triangles using Pythagoras' theorem and the trigonometric ratios, including angles of elevation and depression and problems involving true and compass bearings”

HSCMaths Standard 1Year 12: Measurement8 min read

Quick answer

Use Pythagoras' theorem for sides and SOH CAH TOA for sides and angles in right-angled triangles. Angles of elevation (up) and depression (down) are measured from the horizontal and are equal by alternate angles. True bearings are three-digit angles clockwise from north; compass bearings start with N or S. Always draw a diagram with north marked.

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Practice questions

What this dot point is asking

You need to solve practical right-angled triangle problems: finding sides with Pythagoras' theorem, finding sides and angles with trigonometric ratios, and applying these to heights and distances (angles of elevation and depression) and navigation (true and compass bearings).

The answer

Pythagoras' theorem and trigonometry

In a right-angled triangle with hypotenuse cc: c2=a2+b2c^2 = a^2 + b^2.

Label sides relative to the angle θ\theta: opposite, adjacent, hypotenuse.

SOH CAH TOA

sin⁡θ=OHcos⁡θ=AHtan⁡θ=OA\sin \theta = \frac{O}{H} \qquad \cos \theta = \frac{A}{H} \qquad \tan \theta = \frac{O}{A}

  • Finding a side: write the ratio, substitute, rearrange.
  • Finding an angle: calculate the ratio, then use the inverse key (sin⁡−1\sin^{-1}, cos⁡−1\cos^{-1}, tan⁡−1\tan^{-1}).
  • Calculator in degree mode.

Angles of elevation and depression

  • Angle of elevation: from the horizontal up to the object.
  • Angle of depression: from the horizontal down to the object.
  • They are equal (alternate angles between parallel horizontal lines), which lets you put the angle inside the triangle.

Bearings

  • True bearings are measured clockwise from north, written with three digits: 045°T, 230°T.
  • Compass bearings use N or S first, then an angle towards E or W: N45°E, S50°W.
  • Always draw a north line at the point you are measuring from.
  • Reverse bearing: add or subtract 180°.
Worked example

A ship sails 12 km on a bearing of 040°T. How far north and how far east of its start is it?

  1. The angle between the path and north is 40°.
  2. North component (adjacent): 12cos⁡40°=9.1912 \cos 40° = 9.19 km.
  3. East component (opposite): 12sin⁡40°=7.7112 \sin 40° = 7.71 km.
  4. The ship is about 9.2 km north and 7.7 km east of its start.
Common traps
Calculator in radians
Check for degree mode.
Measuring bearings from the wrong point
"The bearing of B from A" is measured at A.
Putting the angle of depression at the wrong corner
Use alternate angles to move it into the triangle.

Practice questions

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

foundation3 marks
A ramp is 6 m long and rises 0.8 m. Find (a) the horizontal distance to 2 decimal places and (b) the angle the ramp makes with the ground to the nearest degree.
Show worked solution →

(a) h2=62−0.82=36−0.64=35.36h^2 = 6^2 - 0.8^2 = 36 - 0.64 = 35.36, so h=5.95h = 5.95 m.

(b) sin⁡θ=0.86=0.1333\sin \theta = \frac{0.8}{6} = 0.1333, so θ=8°\theta = 8° (7.66° rounded).

Marking guide: 1 mark for (a), 2 marks for (b).

core4 marks
From the top of a 45 m cliff, the angle of depression to a boat is 18°. How far is the boat from the base of the cliff, to the nearest metre?
Show worked solution →

The angle of depression from the cliff top equals the angle of elevation from the boat (alternate angles), so the angle at the boat is 18°.

tan⁡18°=45d\tan 18° = \frac{45}{d}, so d=45tan⁡18°=138.49d = \frac{45}{\tan 18°} = 138.49 m.

The boat is 138 m from the base of the cliff (to the nearest metre).

Marking guide: 1 mark for a correct diagram, 1 mark for the correct ratio, 1 mark for rearranging, 1 mark for the answer.

exam5 marks
A hiker walks 5 km due east from camp, then 3 km due south. (a) How far is the hiker from camp? (b) Find the true bearing of the hiker from camp, to the nearest degree.
Show worked solution →

(a) d2=52+32=34d^2 = 5^2 + 3^2 = 34, so d=5.83d = 5.83 km.

(b) At camp, the angle south of east is θ\theta where tan⁡θ=35\tan \theta = \frac{3}{5}, so θ=31°\theta = 31°.

The bearing is measured clockwise from north: east is 90°, then a further 31° towards south, so the bearing is 90°+31°=90° + 31° = 121°T.

Marking guide: 2 marks for (a), 1 mark for the angle, 1 mark for the diagram showing north, 1 mark for 121°T.

Practise this

Sources & how we know this

Based on the official NESA syllabus. Not yet individually reviewed by a person; always check NESA and your school's materials for current rules.

ExamExplained