Three-Dimensional Trigonometry (Grade 12)

Applying the sine, cosine and area rules to problems set in three dimensions — identifying the right triangle to work in, finding the shared side first, and then answering the question.

A 3-D question is really two 2-D triangles joined by a shared side. Find that shared side in one triangle, then carry it into the other. Nothing new is needed beyond the sine, cosine and area rules.

1The method

Identify the horizontal triangle and the vertical one.
Find the side they share — this is almost always the first thing to calculate.
Work in the horizontal triangle first (it usually has the most information).
Carry the shared side into the vertical triangle and finish with basic SOHCAHTOA.

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Angles of elevation are measured from the horizontal. If a pole is vertical, the triangle containing it has a 90° at its base — so that triangle is right-angled and SOHCAHTOA works there.

AB is a vertical pole. B, C and D lie in the same horizontal plane, with BD̂C = 40°, BĈD = 65° and CD = 30 m. The angle of elevation of A from C is 32°.

Worked ExampleCalculate the height of the pole AB in the diagram above

1
Work in the horizontal triangle BCD first.
It has two angles and a side — enough for the sine rule.
2
DB̂C = 180° − 40° − 65° = 75°.
Angles of a triangle.
3
BCsin 40° = 30sin 75°.
BC is opposite the 40°; CD (=30) is opposite the 75°.
4
BC = 30 sin 40°sin 75° = 19,96 m.
This is the shared side.
5
Now the vertical triangle ABC: AB̂C = 90° because AB is vertical.
The pole meets the horizontal plane at a right angle.
6
tan 32° = ABBC, so AB = 19,96 × tan 32°.
SOHCAHTOA — the elevation angle is at C.
7
AB = 12,47 m.
Shorter than BC, which is right for a 32° elevation. Sensible.

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Do not try to apply the sine rule to the whole 3-D figure at once. It only works inside a single plane triangle. Split the problem into the horizontal one and the vertical one.

2When the shared side is unknown in both triangles

If neither triangle can be solved alone, let the shared side be x, write it two ways, and set them equal. This produces the general expressions that Paper 2 asks you to 'show that'.

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In a 'show that' question, work towards the printed answer and stop when you reach it. Do not carry on and produce a decimal — the expression is the answer.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
In the diagram below, calculate the size of DB̂C.
Show answer ▾Hide answer ▴
180° − 40° − 65° = 75°
2
In the diagram below, calculate the length of BC.
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BCsin40° = 30sin75° → BC = 19,96 m
3
In the diagram below, calculate the height of the pole AB.
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AB = BC tan32° = 19,96 × tan32° = 12,47 m
4
Why is AB̂C = 90° in the diagram below?
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Because AB is vertical and BC lies in the horizontal plane.
5
Which triangle should you solve first in a 3-D problem?
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The one with the most information — usually the horizontal triangle — so that you can find the shared side.
6
An angle of elevation is measured from which direction?
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From the horizontal.
7
Calculate the area of △BCD given BC = 19,96 m, CD = 30 m and BĈD = 65°.
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12(19,96)(30)sin65° = 271,40 m²
8
Can the sine rule be applied to a three-dimensional figure as a whole?
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No — it applies only within a single plane triangle, so the figure must be split up.