Mathematics, four units · Grade 11

How do you solve a trigonometry question set in a trapezium or a pyramid?

Find a triangle inside it. The examined question is almost never a bare triangle, so the first move is to draw a height or a diagonal that breaks the shape into triangles. Once there is a triangle with enough data in it, you carry on with the sine or cosine rule exactly as usual.

Learning objectives

The construction line is the solution

Nearly every question about a quadrilateral opens with the same move: add one line that breaks the shape into triangles. In a trapezium drop a height from one vertex or from both; in a parallelogram draw a diagonal.

The choice is not arbitrary. Look at which data you already have and draw the line that produces a triangle containing most of them together. A construction line that scatters the data across two triangles does not help.

After drawing, write down explicitly which triangle you are working in and what is known there. This is not a formality — an examiner who cannot tell which triangle you used cannot award the mark for the step.

Trapezia and parallelograms

In a right-angled trapezium the height equals the perpendicular leg, and the longer base splits into the shorter base plus a remaining segment. That segment is the horizontal side of the triangle you created.

In an isosceles trapezium, heights dropped from the two upper vertices cut off equal pieces at both ends. The middle piece equals the shorter base and the two overhangs are equal to each other.

In a parallelogram a diagonal splits it into two congruent triangles. A given angle also gives you its supplement to 180 degrees at the adjacent vertex, and that is often precisely the missing fact.

Solids: the angle with the base

In a pyramid, the angle between a lateral edge and the base is measured between that edge and its projection onto the base — the segment joining the centre of the base to the base vertex.

So the working triangle is the right-angled one whose vertices are the apex, the centre of the base and a base vertex. The height of the pyramid is one leg and half the base diagonal is the other.

The angle between a lateral face and the base is a different question. There the projection runs from the midpoint of a base edge rather than from a vertex, so the second leg is half the edge. Confusing the two is the central error in solids.

Worked examples

  1. A right-angled trapezium has bases 12 and 8 and a perpendicular leg of 5. Find the slanted leg.

    1. Drop a height from the far upper vertex onto the longer base
    2. This gives a right-angled triangle with legs 5 and 12 minus 8, that is 4
    3. By Pythagoras: 25 plus 16

    Answer: The square root of 41, about 6.4

  2. In the same trapezium, what is the angle between the slanted leg and the longer base?

    1. In that right-angled triangle, the leg opposite the angle is 5
    2. The leg adjacent to it is 4
    3. The tangent of the angle is 5 over 4

    Answer: About 51.3 degrees

  3. A pyramid on a square base of side 6 has height 4. Find the angle between a lateral edge and the base.

    1. Base diagonal: 6 times the square root of 2, about 8.49
    2. Half the diagonal is about 4.24
    3. In the right-angled triangle: height 4 against half-diagonal 4.24
    4. The tangent of the angle is 4 over 4.24

    Answer: About 43.3 degrees

Common mistakes

Trying to solve a quadrilateral without a construction line
There is no quadrilateral version of the sine or cosine rule. Without a height or a diagonal there is no triangle, and no theorem that can be applied to the data.
Confusing the edge angle with the face angle
The edge angle is measured to half the diagonal and the face angle to half the side. They differ in size, and using one for the other is wrong by several degrees.
Not stating which triangle the calculation is in
On a diagram with three triangles there is no way to tell what was computed. Naming the triangle at each step is part of the justification and is marked as such.

What to remember

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