Formulae for the calculation of triangles

In order to compute elements of an arbitrary triangle, you can decompose it by means of heights into right-angled triangles and then deal with these one by one. It will be better not to follow this laborious task over and over again, but to use it to derive generally valid formulae. For example, if you express in the second triangle above h_c once in terms of the items of the triangle DBC, a second time in terms of the items of the triangle ADC, you should be able to find relations between the angles and sides of the arbitrary triangle ABC.

Sin Rule

First proof: Referring to the second triangle above, you find

sin = h_c/b, whence h_c = b sin and sin = h_c/a. whence h_c = a sin .

Eliminating h_c, you obtain

b sin = a sin

or

1) b/sin = a/sin .

The last figure above yields

sin = h_a/b, whence h_a = b sin and sin = h_a/c. whence h_a = c sin .

Eliminating h_a, you obtain

b sin = c sin

or

2) b/sin = c/sin .

The results 1) and 2) yield the Sin Rule: a/sin = b/sin = c/sin , which was first reported by the Persian astronomer Nasis Eddin Tusi (1201 - 1274).

Second proof: Draw about the triangle ABC the circumscribed circle; then AMB as central angle is twice as large as the periphery angle ACB over the arc. The triangle AMB is isosceles, whence DMB = and sin =(c/2)/ r, whence c/sin = 2r. If you had started from the triangle MBC or AMC, you would have found correspondingly the Sin Theorem:

a/sin a = b/sin = c/sin = 2r.

In right-angled and isosceles triangles, only sharp angles were possible, but now you can also have obtuse angles. What form does the Sin Rule then take? Let be an obtuse angle, so that h_c lies outside the triangle ABC. You have now

sin = h_c/a, sin (180º - a ) = = h_c/b,

whence sin(180º- a )=sin a , asinb=b or a/b = sin a /sin b, etc.

We must still look at the case when, for example, g= 90º. Since sin 90º = 1, a/sin a = c/ 1 or sin a = a/c, b/ sin b = c/1 or sin b = b/c. The Sin Theorem and the definition of sin become identical for right-angled triangles.

Cos Theorem

You have sin a = h_c/b or h_c = bsin a , cos a = q/b or q = bcos a . Applying Pythagoras' theorem to the first triangle below, then a²=h_c²+(c-bcosa)², or, substituting the expressions above:

a² = (bsin a)² + (c - bcosa)² = b²sin²a + c² - 2bc cosa + b²cos²a =
= b²(sin²a+ cos²a) + c - 2bc cosa,

and, since sin²a+cos²a = 1,

a² = b² + c² - 2bc cos a.

In the same manner, you obtain the equations for b² and c², which follow also more simply by changing the notation in the original triangle, that is,

Cos rule

a² = b² + c² - 2bc cos a, b² = a² + c² - 2ac cos b, c² = a² + b² - 2ab cos g.

It was first reported by the French mathematician Vieta.

You obtain out of one formula the others by cyclic interchange, that is, by substitution of the symbols of the sides clockwise and those of the angles anti-clockwise, as shown. If an angle is obtuse, for example a, then h_c= bsin (180º-a) =b sina; qbcos(180º - a)=-b cos a and

a² = h_c² + (c+q)² = (b sin a)² + (c - b cos a)². You find for a² the same expression as before, and therefore the same final result.

Finally, if one angle is a right one, for example , then, since cos 90º = 0, c²=a²+b²-2abcosg becomes c²=a²+b² - Pythagoras' Theorem - and moreover

in the definition of the cos. The Cos Rule is also referred to as the extended Pythagorean Theorem.

Area formulae

In particular, for the right-angled triangle with sides a and b, F = ab/2. Since in tasks dealing with triangles the lengths of heights are not at all computed, it will be of use to express the area directly in terms of sides and angles.

Since sin a =h_c/b or h_c=bsin a and therefore F=c·h_c/2=c·bsina/2, you repeat these manipulations for h_a and h_b:

F = ab sin g/2 = bc sin a/2 = ac sin b/2.

During land surveys, you are dealing mostly with large distances, the exact measurement of which is difficult. For this reason, as far as it is possible, you measure angles rather than distances and reduce areas to angles.

The Sin Rule yields: b=asin b/sin a, which, on substitution into F=absing/2, becomes:

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