Solve right-triangle and bearing problems
Choose sine, cosine or tangent from the sides relative to an acute angle; model two-dimensional heights, bearings and shortest distances.
Section 7 · Trigonometry
Use this pathway for Cambridge IGCSE International Mathematics 0607, 2028–2030. Core and Extended objectives are labelled in each lesson; Extended adds requirements beyond Core. Paper 1 and Paper 2 are non-calculator. Papers 3–6 require a graphic display calculator (GDC). These original lessons use self-checked practice; they are not a Cambridge grade, full timed paper or teacher review.
Before you begin
- Check whether your school has entered you for Core or Extended. Core and Extended share most Number ideas, but some notation and methods are Extended only.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which three sides define the right-triangle ratios?
What you will learn
- Select a trigonometric ratio for a missing side or acute angle
- Solve two-dimensional problems with Pythagoras and trigonometry
- Interpret bearings, elevation and depression and perpendicular distance
Label the sides from the chosen angle
SOH: sin θ = opposite/hypotenuse; CAH: cos θ = adjacent/hypotenuse; TOA: tan θ = opposite/adjacent. The hypotenuse is opposite 90°; opposite and adjacent depend on the chosen acute angle.
Use inverse sin, cos or tan for an angle. Work in degree mode and round angles to one decimal place as required. Keep unrounded values until the final step.
Draw and label the situation
For elevation or depression, add a horizontal eye-level line and the vertical height. A bearing is measured clockwise from north and written with three digits; resolve the direction into a right triangle before calculating.
The shortest distance from a point to a straight line is the perpendicular distance. In a diagram this is the 90° segment to the line. Check that the sought angle or distance is measured from the correct side.
Pause and explain
From which direction and in which direction is a bearing measured?
Worked example
A 7 m ladder rests against a vertical wall. Its foot is 2 m from the wall. Find the angle between the ladder and the ground, to 1 decimal place.
Show the worked solution
- The ladder is the hypotenuse and the ground distance is adjacent to the angle.
- cos θ = 2/7.
- θ = cos⁻¹(2/7) ≈ 73.4°.
Answer 73.4°.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A ramp rises 1.8 m over a horizontal run of 2.4 m. Find its angle above the ground to 1 decimal place.
Give me a hint
The height is opposite and the run is adjacent to the angle.
Compare my reasoning
- tan θ = 1.8/2.4 = 0.75.
- θ = tan⁻¹(0.75).
- θ ≈ 36.9°.
36.9°.
Look for these in your work
- Opposite and adjacent identified
- Tangent selected
- Degree answer rounded to 1 d.p.
From a point on level ground, the angle of elevation to a tower top is 32°. The point is 40 m from the tower base. Find the tower height to 1 decimal place.
Give me a hint
The horizontal distance is adjacent; height is opposite.
Compare my reasoning
- tan 32° = height/40.
- height = 40 tan 32°.
- height ≈ 25.0 m.
25.0 m.
Look for these in your work
- Diagram interpreted
- Tangent relation set up
- Height unit stated
A rescue point is 30 m east and 40 m north of a lookout. Find the distance and the bearing from the lookout to the point.
Give me a hint
Use a right triangle for distance and measure the acute direction clockwise from north.
Compare my reasoning
- Distance = √(30² + 40²) = 50 m.
- The angle east of north is tan⁻¹(30/40) ≈ 36.9°.
- Bearing = 036.9°, or 037° to the nearest degree.
Distance 50 m; bearing 037° to the nearest degree.
Look for these in your work
- North reference used
- Distance calculated
- Three-digit bearing reported
Common mistakes
- Swapping opposite and adjacent after changing the reference angle
- Using a bearing measured anticlockwise from north
- Rounding intermediate values
- Treating a sloping sightline as the vertical height
What can you explain now?
Relative to angle θ, opposite = 5 and adjacent = 12. Which ratio gives θ?
Compare with the explanation
tan θ = 5/12
Tangent is opposite divided by adjacent.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
How do the chosen acute angle and diagram determine whether to use sine, cosine or tangent?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.