Calculation of arbitrary triangles

Four basic tasks

With the aid of the preceding formulae, you can now solve the four fundamental tasks which arise in design theory .

First basic task: Analyze a triangle, given the two sides a, b and the angle between them.

Solution:

As long as < 180º, this task has only one solution.

Second basic task: Analyze a triangle, given one side a and two angles a, b.

Solution:

a.		g = 180º - (a + b)
b.		a/sina = b/sin b		b = (a/sina) · sin b
c.		a/sin =c/sin g		c= (a/sin) ·sin g

As long as a+b < 180º, this task has one and only one solution.

Third basic task: Analyze a triangle, given the three side a, b, c.

Solution:

As long as the sum of any two sides is larger than the third, this task has only one solution.

Fourth basic task: Analyze a triangle, given the two sides a, b (a > b) and the angle a opposite to the larger side.

Solution:

As long as < 180º, this task has only one solution.

If, given two angles, you include the third angle among the given items, you have the simple rules:

Rule 1 for the basic tasks :

If you have among the given items an angle and its opposite side,
then the Sine Rule yields the answer, otherwise the Cosine Rule.

There still exists the possibility that the triangle under consideration has an obtuse angle. In the case of cos a, there arises no difficulty. If cos a > 0, a is a sharp angle, if cos a < 0, a is obtuse, since angles cannot exceed 180º. It is different, if the angle is to be dermined from sin a, since with a positive value, a can be sharp or obtuse. Since the smaller angle lies opposite the smaller side, you can avoid unnecessary extensions, if you compute first the angle opposite the smaller side, which must always be sharp. Only the angle opposite the longest side can be obtuse! Therefore it is best to determine the largest angle by subtracting the other two angles from 180º.

Rule 2 for the basic tasks:

cos yields an angle uniquely. When using the sin Rule, take into consideration that opposite the smaller angle lies always a sharp angle!

Given two sides and the side opposite the smaller angle, say, a, b,b, you do not face any more a basic task and the following three cases may arise:

The trigonometric calculation is the same. The sin Rule yields for the given items:

In the case c), you must let a be sharp and obtuse and compute both triangles.

Logarithmic computations

During computations of triangles, you often try to abbreviate the calculations and use a slide rule. When you employ logarithms, it is important to set out the calculations clearly. The following scheme has its logarithmic part follow the pattern of the tables:

Task 1: Given c = 36.74 m, a = 45.12º and b = 26.64º, compute the missing siedes and angles of the triangle.

Task 2: Given a = 96.14 m, b = 41.43 m and g = 30.17º, compute the other angles and side of the triangle.!

Method of partial triangles

If you are not given directly one of the four fundamental tasks, as during design tasks, partial triangles can be used. Once one of them has been solved, you can proceed to another one, etc. until the required items have been determined.

The geometrical properties of a figure
must always be taken into account.

Task 1. Given , e and a, find b!

The partial triangle ECB can be computed using a. CB = a, b. CEB = e and c. CE = , according to the Basic Task IV. Therefore you can assume that all items are known. Now there arises the question which item to find, in order to compute another triangle. Our goal is b, which occurs in the triangles ACE and ABC. You can use both triangles and arrive at the sequence of computations:

EBC:

a/sin e = /sin b

whence

sin b =·sin e/a

and

g / 2 = 180º - (e + )

Once g and b are known, a is also known. Now you proceed to the triangle ACE or ACB, each of which can be analyzed according to the Basic task II:

ACE		b/sin(180º - e) = /sinb	whence	b = ·sin e /sina
ABC		b/sin b = a/sin a	"	b = asin b /sin a

Although the two expressions for b differ, they yield nevertheless the same result, since

sin b = sine / a yields sin e = asin b.

Task 5: Find the height of an object (tower, etc.).

You have measured the base line AB = s and the angles a, b and g . The partial triangle ABF can be computed according to the Basic task II, whence AF can be found. Then h is found from the right-angled triangle AFS.

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