Since 
(see
Problem
19, Exercises for Chapter 3 Section 1.) 
If 
is acute
as in Figure 3a, then 
If is obtuse
as in Figure 3b, then 
and
In either
case, 
Equating
these values of
|
|
|
|
|
|
4. The Laws of Sines and Cosines.
Consider the triangles in Figure 3a and Figure 3b. Since every triangle
has at least two acute angles, we can assume we have picked one of them
to call
Since (see
Problem
19, Exercises for Chapter 3 Section 1.)
If is acute
as in Figure 3a, then
If is obtuse
as in Figure 3b, then
and In either
case, Equating
these values of
Figure 3a |
Figure 3b |
h, we have 
or Similarly,
if  is the
angle at C and c is its opposite side, we can show that
We are now prepared to state the Law of Sines: If a triangle
has angles with
opposite sides a, b, and c respectively, (See Figure 3c)
then |
Figure 3c |
All of the examples and problems which follow are based on triangles labeled as in Figure 3c.
There are three important congruence theorems from geometry which are usually abbreviated as ASA, SAS, and SSS. ASA, for example, tells us that if two angles and the included side of one triangle are congruent to the respective two angles and included side of another triangle, the two triangles are congruent. Similar statements apply to the other two abbreviations. What these theorems tell us, is that if the right three parts of a triangle are known, the other three parts are fixed and they should be able to be found. Having partial information about a triangle and using it to find the rest of the information about the triangle is referred to as solving the triangle.
In 
and
c
= 47.26. Solve the triangle. Find the angles to one decimal place and the
lengths to two decimal places.
Solution: First 
so on the calculator we key 180 ENTER 37.1 - 58.3 - and see the answer 
From the law of sines, 
We solve this for a and substitute the known quantities to get 
With the calculator in degree mode and set to Fix 2, we key in 47.26 ENTER
37.1 SIN 
84.6
SIN 
and
obtain
a = 28.63. Similarly 
Figure 4
Consider the complex numbers P and Q as shown in Figure 4. By Problem 33 of Exercises for Chapter II Sections 4, 5 and 6, we see that d, the distance between P and Q, is
But 
Substituting these into the equation for d, squaring both sides,
expanding, and collecting like terms, gives us
Using the Pythagorean Identity [See Problem 7, Exercise for Chapter III Section 1.] and the subtraction formula for the cosine [See Problem 13, Exercise for Chapter III Section 1.] this becomes
If 
is now relabeled as in Figure 3c, we get the three following forms of the
Law
of Cosines
depending on whether angle A, B or C, respectively is placed at the origin.
Given that in 
one has 
a = 22.16, and b = 43.26, solve the triangle. Give angles to
three decimal places and lengths to 2 decimal places.
Solution: We observe that the given information is of the form SAS, hence the triangle is determined. We first use the Law of Cosines to find
On the calculator (in radian mode) this is accomplished with 22.16 LS
x2 43.26 LS x2 + 2 ENTER 22.16
43.26 
COS 
which gives us c = 26.50. We now use the Law of Sines to find
or
Assuming the value of c is still on the stack from the previous
calculation, we proceed with 22.16 ENTER 
SIN 
LS ASIN,
which gives us 
( Notice that when we were ready to divide by c we took advantage
of the fact that it was already on the stack and just used the SWAP command
to put it in the right position for the division. Every time one can eliminate
the need to key in a number, one has removed an opportunity to make an
error.) Finally, we compute the last angle by subtracting the first two
from
and get 
|
|
|
|
|
|