2D and 3D Trigonometry Applications
Key Terms
- In 2D problems: extract the triangle from the context, label all known sides and angles, then choose sine or cosine rule
- Bearings
- are measured clockwise from north (000° to 360°); convert to interior angles before applying trig
- In 3D problems: identify the right triangle or non-right triangle that contains the required side or angle
- Angle of elevation
- : angle measured upward from horizontal; depression: downward from horizontal
- Strategy: draw a 3D diagram, then extract 2D triangles and solve each in sequence
Cosine Rule (side): c² = a² + b² − 2ab cos C
Cosine Rule (angle): cos C = (a² + b² − c²)/(2ab)
Area: Area = ½ab sin C
Right-triangle: tan θ = opp/adj, sin θ = opp/hyp, cos θ = adj/hyp
3D Approach:
• Find horizontal distance first (base triangle), then use elevation/depression angle
• Space diagonal of a rectangular box: d = √(l² + w² + h²)
Step 1: The angle at A between PA and AB: bearing changes from 060° to 150°, so the interior angle of the triangle at A = 150° − 60° = 90°. Wait — the bearing of AB is 150° and back-bearing of AP is 060° + 180° = 240°, so the turn angle at A = 240° − 150° = 90°.
Step 2: Triangle PAB has a right angle at A: PA = 8, AB = 5.
PB = √(8² + 5²) = √89 ≈ 9.43 km
Step 1: Find the diagonal of the rectangular base: AD = √(40² + 30²) = √(1600 + 900) = √2500 = 50 m.
Step 2: Triangle TAD is right-angled at A, with TA = 30 m (vertical), AD = 50 m (horizontal).
tan(θ) = 30/50 = 0.6
θ = tan−1(0.6) ≈ 31.0°
Multi-Step Trigonometry: The Big Picture
In earlier lessons, the sine and cosine rules were applied to single triangles. Real-world problems, however, often involve sequences of triangles — sometimes two or three triangles linked together, with the answer from one feeding into the next. This lesson develops the systematic skill of breaking complex geometric situations into manageable sub-problems, each solved by the appropriate trig rule.
The guiding principle is always the same: draw a clear, labelled diagram, identify every known quantity, and work forward one triangle at a time.
2D Problems: Bearings and Navigation
Bearing problems describe directions using three-digit bearings measured clockwise from north. When two legs of a journey are described by bearings, the key challenge is finding the interior angle of the triangle formed by the starting point, the turning point, and the final point.
Finding the interior angle: The interior angle at the turning point equals the difference between the back-bearing of the first leg and the bearing of the second leg. The back-bearing of a direction is obtained by adding (or subtracting) 180°. Alternatively, draw north lines at each point and use co-interior or alternate angle relationships.
Once the interior angle is found and the triangle has two sides and an included angle (SAS), the cosine rule gives the third side directly. If two angles and a side are known (AAS), use the sine rule.
Area of Triangles from Bearings
Once a triangle is fully defined, its area can be computed using Area = ½ab sin C, where C is the interior angle between sides a and b. In navigation problems, this might represent the area of a region bounded by two travel paths and the direct return route.
3D Problems: Reducing to 2D
Three-dimensional trigonometry problems are solved by reducing them to sequences of 2D triangles. A 3D shape — such as a rectangular prism, pyramid, or wedge — contains many 2D cross-sections that are right triangles or non-right triangles. The skill is identifying which 2D triangle contains the unknown you need.
The standard process for 3D problems:
- Draw a 3D sketch and label all given dimensions.
- Identify the unknown — it will lie in some plane (often horizontal or vertical).
- Find the relevant 2D triangle by slicing the 3D shape along that plane.
- Solve the 2D triangle, potentially feeding results into a second 2D triangle.
Angle of Elevation and Depression
The angle of elevation from point A to point B is the angle between the horizontal at A and the line AB, measured upward. The angle of depression is measured downward from the horizontal. Both angles are always acute and are equal by alternate angles when A and B are at different heights (the elevation from below equals the depression from above, looking at the same line).
In 3D problems involving elevation or depression, the horizontal distance is the base of the right triangle and must be found first — typically by solving the base triangle of the 3D shape.
Space Diagonal of a Box
The space diagonal d of a rectangular box with length l, width w, and height h is found by applying Pythagoras twice:
d = √(l² + w² + h²)
This result arises by first finding the diagonal of the base (√(l² + w²)), then treating that as one leg of a right triangle with height h as the other leg. The angle the space diagonal makes with the base is then θ = tan−1(h / √(l² + w²)).
Angles in Pyramids
For a right pyramid (apex directly above the centre of the base), the angle between a slant edge and the base involves: (1) find the distance from the centre of the base to a base vertex (using base dimensions), then (2) use that distance as adjacent and the height as opposite in a right triangle.
The slant height (distance from apex to the midpoint of a base edge) and the slant edge (from apex to a base vertex) are different — always identify which is being asked for.
Mastery Practice
- A ship travels 12 km due north from port P to point A, then 9 km due east from A to point B. Find the straight-line distance from P to B and the bearing of B from P.
- Find the angle of elevation of the top of a 40 m building from a point on the ground 60 m from its base.
- A rectangular box has length 8 m, width 6 m, and height 3 m. Find the length of the space diagonal.
- Two points A and B are on the same side of a building. From A, the angle of elevation of the top T is 50°. From B (20 m further away along the same line), the angle of elevation is 35°. Find the height of the building.
- A yacht sails 15 km on a bearing of 040° from port P to point A. It then sails 10 km on a bearing of 130° from A to point B. Find the distance PB and the bearing of B from P.
- A right pyramid has a square base of side 10 m and a vertical height of 12 m. Find (a) the slant height (from apex to midpoint of a base edge), and (b) the angle the slant edge makes with the base.
- From the top of a cliff 80 m high, the angles of depression of two boats directly out to sea are 25° and 40°. Find the distance between the two boats.
- Two towns A and B are 50 km apart. Town C is on a bearing of 070° from A and on a bearing of 310° from B. Find the distances AC and BC.
- A rectangular field ABCD has AB = 80 m and BC = 60 m. A vertical pole TP stands at corner A, where T is the top of the pole. From corner C, the angle of elevation of T is 20°. Find the height of the pole, the distance CT, and the angle of elevation of T from the midpoint M of BC.
- Three surveying points A, B, C lie in a horizontal plane. AB = 120 m, AC = 150 m, and the angle BAC = 55°. A vertical tower of height 60 m stands at C. Find the area of triangle ABC and the angle of elevation of the top of the tower from point B.