Learning Objectives
- Solve bearings, elevation and depression problems in two dimensions.
- Apply Pythagoras and trigonometry in three-dimensional solids.
- Calculate the angle between a line and a plane.
- Use scale diagrams and clear sketches to identify right triangles.
- Combine multiple stages without premature rounding.
Key Terms
- Angle of elevation
- Angle measured upward from a horizontal line of sight.
- Angle of depression
- Angle measured downward from a horizontal line of sight.
- Three-figure bearing
- A clockwise angle from north written with three digits.
- Projection
- The shadow or corresponding line of a spatial segment on a plane.
- Line and plane angle
- The angle between a line and its perpendicular projection on the plane.
- Space diagonal
- A diagonal joining opposite vertices through a solid.
What This Chapter Covers
- Bearings and navigation
- Angles of elevation and depression
- Right-triangle modelling
- Pythagoras and trigonometry in 3D
- Angles between lines and planes
Modelling Before Calculating
Application questions are mainly about selecting the correct triangle. Draw a clear sketch, mark horizontal and vertical lines, label north directions in bearing questions, and identify the right angle. The diagram does not need to be to scale, but it must preserve the relationships described.
Bearings
Bearings are measured clockwise from north and written with three digits, such as 045^\circ or 270^\circ. The reverse bearing differs by 180^\circ: add 180 if the original is below 180, otherwise subtract 180.
Worked Example: Reverse Bearing
Question: The bearing of B from A is 068^\circ. Find the bearing of A from B.
- Add 180^\circ because 68 is below 180.
- 68+180=248.
- Write the result with three figures.
Answer: 248^\circ.
Navigation With The Sine And Cosine Rules
Worked Example: Two-Leg Journey
Question: A ship travels 18 km from P on a bearing 040^\circ, then 25 km on a bearing 130^\circ. Find its distance from P.
- The change in bearing is 130-40=90^\circ, so the two legs are perpendicular.
- Use Pythagoras: d=\sqrt{18^2+25^2}.
Answer: d\approx30.8 km.
If the included angle is not 90°, use the cosine rule. Draw a north line at each relevant point because bearings are always measured from a local north direction.
Elevation And Depression
Horizontal lines at different heights are parallel. Therefore an angle of depression from an observer is equal to the corresponding angle of elevation from the ground point. Height is usually opposite the angle and horizontal distance adjacent to it, so tangent is often appropriate.
Worked Example: Building Height
Question: From a point 36 m from the base of a vertical building, the angle of elevation to the top is 52^\circ. Find the building height.
- Use \tan52^\circ=h/36.
- Rearrange: h=36\tan52^\circ.
Answer: h\approx46.1 m.
Two Observation Points
When observations are made from two positions, create one equation from each right triangle. If the points are on the same straight horizontal line, express one distance in terms of the other and the known separation.
Worked Example: Two Angles Of Elevation
Question: Two points A and B lie on the same side of a tower, with B 20 m closer to the tower. The angles of elevation are 30^\circ at A and 45^\circ at B. Find the tower height.
- Let distance from B to the tower be x, so distance from A is x+20.
- From B: h=x\tan45^\circ=x.
- From A: h=(x+20)\tan30^\circ.
- Equate and solve: x=(x+20)/\sqrt3.
Answer: h=x=10(\sqrt3+1)\approx27.3 m.
Pythagoras In Three Dimensions
A common strategy is two-stage Pythagoras: first find a diagonal on one rectangular face, then combine it with the perpendicular third dimension to find a space diagonal.
Space diagonal of a cuboid with length l, width w and height h.
Worked Example: Cuboid Diagonal
Question: A cuboid measures 8 cm by 6 cm by 12 cm. Find its space diagonal.
- Base diagonal =\sqrt{8^2+6^2}=10.
- Space diagonal =\sqrt{10^2+12^2}=\sqrt{244}.
Answer: \approx15.6 cm.
Angle Between A Line And A Plane
The required angle is between the line and its perpendicular projection on the plane. Identify the foot of the perpendicular, draw the projection and use the right triangle formed by the line, its projection and the perpendicular height.
Worked Example: Line–Plane Angle
Question: In the cuboid above, find the angle between the space diagonal and the base plane.
- The projection of the space diagonal on the base is the base diagonal, length 10.
- The vertical height is 12.
- If the angle with the base is \theta, \tan\theta=12/10.
Answer: \theta\approx50.2^\circ.
Accuracy In Multi-Stage Problems
Keep unrounded calculator values for intermediate lengths. If a rounded value is used in a later trigonometric calculation, the final answer may lose accuracy. Cambridge expects non-exact numerical answers to three significant figures and angles to one decimal place unless the question states otherwise.
Examination Guidance
- Draw north lines and mark bearings clockwise.
- Use the reverse-bearing rule carefully.
- For elevation and depression, mark the horizontal line of sight.
- In 3D, locate a right triangle and its projection before selecting a trigonometric ratio.
- Keep full calculator values until the final answer.
Common Mistakes
- Measuring a bearing anticlockwise.
- Writing a bearing with fewer than three digits.
- Using the line and the vertical edge instead of the line’s projection when finding an angle with a plane.
- Assuming a 3D diagram is drawn to scale.
- Rounding an intermediate diagonal before using it again.
Knowledge Check And Practice
1. Find the reverse bearing of 215^\circ.
2. A tree is 25 m away and has angle of elevation 40^\circ. Find its height, ignoring eye level.
3. Find the space diagonal of a cuboid 3 by 4 by 12.
4. How is an angle of depression measured?
5. What is a three-figure bearing of due west?
6. Why is the projection needed for a line–plane angle?
