5.7: The Sine Rule and the Cosine Rule
Where given information, or a specified geometrical construction, determines an angle or length uniquely, it is sometimes - but not always - possible to find this angle or length using simple-minded angle-chasing and congruence.
Problem 191
(a) In the quadrilateral ABCD the two diagonals AC and BD cross at X . Suppose AB = BC , ∠ B AC = 60°, ∠DAC = 40°, ∠ B XC = 100°.
(i) Calculate (exactly) ∠ADB and ∠ C BD.
(ii) Calculate ∠ B DC and ∠ACD.
(b) In the quadrilateral ABCD the two diagonals AC and BD cross at X . Suppose AB = BC , ∠ B AC = 70°, ∠DAC = 40°, ∠ B XC = 100°.
(i) Calculate (exactly) the size of ∠ B DC + ∠ACD.
(ii) Explain how we can be sure that ∠ B DC and ∠ACD are uniquely determined, even though we cannot calculate them immediately.
If it turns out that the simplest tools do not allow us to determine angles and lengths, this is usually because we are only using the most basic properties: the congruence criteria, and the parallel criterion. The general art of ‘solving triangles’ depends on the similarity criterion (usually via trigonometry). And the two standard techniques for ‘solving triangles’ that go beyond “angle-chasing” and congruence are the
Sine Rule,
which was established back in Problem
32
(and its consequences, such as the area formula
The next problem invites you to use Pythagoras’ Theorem to prove the Cosine Rule - an extension of Pythagoras’ Theorem which applies to all triangles ABC (including those where the angle at C may not be a right angle).
Problem 192 (The Cosine Rule) Given Δ ABC , let the perpendicular from A to BC meet BC at P . If P = C, then we know (by Pythagoras’ Theorem) that c 2 = a 2 + b 2 . Suppose .
(i) Suppose first that P lies on the line segment CB , or on CB extended beyond B. Express the lengths of PC and AP in terms of b and ∠ C . Then apply Pythagoras’ Theorem to A ΔAPB to conclude that
(ii) Suppose next that P lies on the line segment BC extended beyond C . Prove once again that
Problem 193 Go back to the configuration in Problem 191 (b). The required angles are unaffected by scaling, so we may choose AB = BC = 1. Devise a strategy using the Sine Rule and the Cosine Rule to calculate ∠ B DC and ∠ACD exactly.
It is worth reflecting on what the Cosine Rule really tells us:
(i) if in a triangle, we know any two sides (a and b) and the included angle (C), then we can calculate the third side (c); and
(ii) if we know all three sides (a, b, c), then we can calculate any angle (say C).
Hence if we know three sides, or two sides and the angle between them, we can work out all of the angles. The Sine Rule then complements this by ensuring that:
(iii) if we know any side and two angles (in which case we also know the third angle), then we can calculate the other two sides; and
(iv) if we know any angle A , and two sides - one of which is the side a opposite A, then we can calculate (one and hence) both the other angles (and hence the third side).
The upshot is that once a triangle is uniquely determined by the given data, we can “solve” to find all three sides and all three angles.
Trigonometry has a long and very interesting history (which is not at all easy to unravel). Euclid (flourished c. 300 BC) understood that corresponding sides in similar figures were “proportional”. And he stated and proved the generalization of Pythagoras’ Theorem, which we now call the Cosine Rule; but he did this in a theoretical form, without introducing cosines. Euclid’s versions for acute-angled and obtuse-angled triangles involved correction terms with opposite signs, so he proved them separately (Elements , Book II, Propositions 12 and 13).
However, the development of trigonometry as an effective theoretical and practical tool seems to have been due to Hipparchus (died c. 125 BC), to Menelaus (c. 70-130 AD), and to Ptolemy (died 168 AD). Once trigonometry moved beyond the purely theoretical, the combination of
- the (exact) language of trigonometry, together with the Sine Rule and the Cosine Rule, and
- (approximate) “tables of trigonometric ratios” (nowadays replaced by calculators)
liberated astronomers, and later engineers, to calculate lengths and angles efficiently, and as accurately as they required.
In mathematics we either work with exact values, or we have to control errors precisely. But trigonometry can still be a valuable exact tool, provided we remember the lessons of working with fractions such as or with surds such as , or with constants such as , and resist the temptation to replace them by some unenlightening approximate decimal. We can replace and by their exact values; but in general we need to be willing to work with, and to think about, exact forms such as “ ” and “ ” , without switching to some approximate evaluation.
Problem 194
(a) Let ABCD be a regular tetrahedron with edges of length 2. Calculate the (exact) angle between the two faces ABC and DBC .
(b) We know that in 2D five equilateral triangles fit together at a point leaving just enough of an angle to allow a sixth triangle to fit. How many identical regular tetrahedra can one fit together, without overlaps around an edge, so that they all share the edge BC (say)?
Problem 195
(a) Let ABCDEF be a regular octahedron with vertices B, C, D, E adjacent to A forming a square BCDE, and with edges of length 2. Calculate the (exact) angle between the two faces ABC and FBC .
(b) How many identical regular octahedra can one fit together around an edge , without overlaps, so that they all share the edge BC (say)?
Problem 196 Go back to the scenario of Problem 188 , with a regular tetrahedron and a regular octahedron both having edges of length 2, and both having one face flat on the table. Suppose we slide the tetrahedron across the table towards the octahedron. What unexpected phenomenon is guaranteed by Problems 194 (a) and 195 (a)?
Problem 197 Consider the cube with edges of length 2 running parallel to the coordinate axes, with its centre at the origin (0,0,0), and with opposite corners at (1,1,1) and (—1, —1, —1). The x- , y- , and z -axes, and the xy- , yz-, and zx -planes cut this cube into eight unit cubes - one sitting in each octant.
(i) Let A = (0,0,1), B = (1, 0, 0), C = (0,1,0), W = (1,1,1). Describe the solid ABCW .
(ii) Let D = (—1, 0, 0), X = (—1,1,1). Describe the solid ACDX .
(iii) Let E = (0, —1,0), Y = (—1, —1,1). Describe the solid ADEY .
(iv) Let Z = (1, —1,1). Describe the solid AEBZ .
(v) Let F = (0,0, —1) and repeat steps (i)-(iv) to obtain the four mirror image solids which lie beneath the xy -plane.
(vi) Describe the solid ABCDEF which is surrounded by the eight identical solids in (i)-(v).
Problem 198 Consider a single face ABCDE of the regular dodecahedron, with edges of length 1, together with the five pentagons adjacent to it - so that each of the vertices A , B , C , D , E has vertex figure 5 3 . Each vertex figure is rigid, so the whole arrangement of six regular pentagons is also rigid. Let V , W , X , Y , Z be the five vertices adjacent to A , B , C , D , E respectively. Calculate the dihedral angle between the two pentagonal faces that meet at the edge AB .
Problem 199 Suppose a regular icosahedron (Problem 189 ) has edges of length 2. Position vertex A at the ‘North pole’, and let BCDEF be the regular pentagon formed by its five neighbours.
(a)(i) Calculate the exact angle between the two faces ABC and ACD.
(ii) How many identical regular icosahedra can one fit together, without overlaps, around a single edge?
(b) Let C be the circumcircle of BCDEF , and let O be the circumcentre of this regular pentagon.
(i) Prove that the three edge lengths of the right-angled triangle ΔBOA are the edge lengths of the regular hexagon inscribed in the circle C , of the regular 10-gon inscribed in the circle C , and of the regular 5-gon inscribed in the circle C .
(ii) Calculate the distance separating the plane of the regular pentagon BCDEF , and the plane of the corresponding regular pentagon joined to the ‘South pole’.
Notice that Problem 199 (b) shows that the regular icosahedron can be ‘constructed’ in the Euclidean spirit: part (b)(i) is essentially Proposition 10 of Book XIII of Euclid’s Elements , and part (b)(ii) is implicit in Proposition 16 of the Book XIII. Once we are given the radius OB , we can:
- construct the regular pentagon BCDEF in the circle C;
- bisect the sides of the regular pentagon and hence construct the regular 10-gon BVC . . . in the same circle;
- construct the vertical perpendicular at O, and transfer the length BV to the point O to determine the vertex A directly above O ;
- transfer the radius OB to the vertical perpendicular at O to determine the plane directly below O , and hence construct the lower regular pentagon; etc..
It may be worth commenting on a common confusion concerning the regular icosahedron. Each regular polyhedron has a circumcentre, with all vertices lying on a corresponding sphere. If we join any triangular face of the regular icosahedron to the circumcentre O , we get a tetrahedron. These 20 tetrahedra are all congruent and fit together exactly at the point O “without gaps or overlaps”. But they are not regular tetrahedra: the circumradius is less than the edge length of the regular icosahedron.
Problem 200 Prove that the only regular polyhedron that tiles 3D (without gaps or overlaps) is the cube.
In one sense the result in Problem 200 is disappointing. However, since we know that there are all sorts of interesting 3-dimensional arrangements related to crystals and the way atoms fit together, the message is really that we need to look beyond regular tilings. For example, the construction in Problem 197 shows how the familiar regular tiling of space with cubes incorporates a semi-regular tiling of space with eight regular tetrahedra and two regular octahedra at each vertex.