LEARNING OBJECTIVES
What you will be able to do
- apply Pythagoras' theorem
- use sine, cosine and tangent in right-angled triangles
- recall exact trigonometric values
- sketch trigonometric graphs and solve equations
- use sine and cosine rules and triangle area
- solve three-dimensional problems
AT A GLANCE
INTRODUCTION · THE BIG IDEA
Connect angles and lengths in right-angled, non-right-angled and three-dimensional problems.
Trigonometry measures triangles by connecting their sides and angles. The method depends on the information given, so begin by marking the known sides, angles and any right angle.
Diagrams organise the geometry; formulas complete the calculation. Keep full calculator values between steps and check that the final length or angle is plausible.
SECTION 01
Pythagoras and right-angled triangles
Pythagoras applies only to right-angled triangles. The hypotenuse is opposite the right angle and is always the longest side. When finding a shorter side, subtract its square from the hypotenuse square.
For trigonometric ratios, label opposite and adjacent relative to the chosen angle. The hypotenuse label never changes. Use inverse sine, cosine or tangent when the unknown is an angle.
Finding an angle in a right-angled triangle
- Opposite side = 8.4 cm and adjacent side = 12.7 cm.
- tan θ = 8.4/12.7.
- θ = tan⁻¹(8.4/12.7).
Answer: θ = 33.5° to one decimal place.
SECTION 02
Elevation, depression and bearings
An angle of elevation is measured upward from a horizontal line; an angle of depression is measured downward. Horizontal lines are parallel, so alternate angles often transfer the given angle into a triangle.
The perpendicular distance from a point to a line is the shortest distance. Bearings may create triangles that need trigonometry.
Height from an angle of elevation
- An observer is 32 m from a vertical tower.
- The angle of elevation to the top is 41°.
- tan 41° = height/32.
- height = 32 tan 41°.
Answer: The tower height is 27.8 m to one decimal place.
SECTION 03
Exact values and trigonometric graphs
Exact values come from special triangles and should be given without decimals. For 0° to 360°, sine and cosine repeat every 360°; tangent repeats every 180°.
Sine and cosine range from −1 to 1. Tangent is undefined at 90° and 270°, creating vertical asymptotes.
Use the graph or reference angle to find every solution in the stated interval.
DETAILED EXPLANATION
- cos values are the sine list in reverse.
- tan 0°, 30°, 45°, 60° = 0, 1/√3, 1, √3.
| x | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin x | 0 | 1/2 | √2/2 | √3/2 | 1 |
| cos x | 1 | √3/2 | √2/2 | 1/2 | 0 |
| tan x | 0 | 1/√3 | 1 | √3 | undefined |
Solving sin x = 0.6 for 0° ≤ x ≤ 360°
- First solution: x = sin⁻¹(0.6) = 36.869…°.
- Sine is positive in quadrants I and II.
- Second solution = 180° − 36.869…°.
Answer: x = 36.9° or 143.1°.
SECTION 04
Non-right-angled triangles
Use the sine rule when a known side is paired with its opposite angle. Use the cosine rule for three sides, or for two sides and the included angle.
The sine rule ambiguous case can produce two possible angles because sin θ = sin(180° − θ). Check whether both form valid triangles.
The triangle area formula needs two sides and their included angle.
DETAILED EXPLANATION
- Example: if sin B = 0.6, test B = 36.9° and B = 143.1° against the other known angle.
- When finding an angle with the cosine rule, rearrange to cos A before using cos⁻¹.
Using the cosine rule
- Two sides are 7 cm and 11 cm with included angle 58°.
- Opposite side c satisfies c² = 7² + 11² − 2(7)(11)cos 58°.
- c² = 88.392…
- Take the positive square root.
Answer: c = 9.40 cm to three significant figures.
SECTION 05
Three-dimensional trigonometry
A 3D problem is usually solved using two connected right-angled triangles. First find a hidden diagonal, then use it in the triangle containing the required angle or length.
The angle between a line and a plane is the angle between the line and its perpendicular projection onto that plane.
Angle between a space diagonal and the base
- A cuboid measures 6 cm by 8 cm by 12 cm.
- The projection of the space diagonal onto the base is the base diagonal √(6² + 8²) = 10 cm.
- In the right triangle, vertical height = 12 cm and adjacent projection = 10 cm.
- tan θ = 12/10, so θ = tan⁻¹(1.2).
Answer: The angle between the space diagonal and the base plane is 50.2°.
QUICK CHAPTER SUMMARY
The ideas to carry forward
- Pythagoras requires a right angle.
- SOHCAHTOA labels depend on the chosen angle.
- Exact values remain in surd or fractional form.
- Trigonometric equations may have several solutions.
- Choose sine or cosine rule from the given information.
- Reduce 3D problems to linked 2D triangles.
QUICK REVISION CHECKLIST
Can you do each of these without your notes?
- apply Pythagoras' theorem
- use sine, cosine and tangent in right-angled triangles
- recall exact trigonometric values
- sketch trigonometric graphs and solve equations
- use sine and cosine rules and triangle area
- solve three-dimensional problems