To solve a triangle means to find all the angles and all the side lengths.
The Law of Sines is a valuable tool in solving triangles,
but it requires knowing an angle and its opposite side.
So, the Law of Sines can't be used as a first step in solving SSS (sidesideside) or SAS (sideangleside) configurations.
In an earlier section, we saw that the Law of Cosines comes to the rescue!
There, an example showed how to use the Law of Cosines in a SAS situation.
Here, we'll talk about using the Law of Cosines in a SSS situation.
To do this, some knowledge of the arccosine is required.
(The arccosine function will be studied in more detail later on—you just get a basic working knowledge here.)
When the Law of Cosines is used in a SSS case (a triangle with sides $\,d\,$, $\,e\,$ and $\,f\,$ known), then the angle $\,D\,$ (across from side $\,d\,$) is the unknown in the following equation: $$ d^2 = e^2 + f^2  2ef\cos D $$ Solving for $\cos D\,$ gives $$ \cos D = \frac{d^2  e^2  f^2}{2ef} $$ The righthand side is just a constant, since $\,d\,$, $\,e\,$ and $\,f\,$ are all known. For the moment, suppose that this righthand side evaluates to (say) $\,0.5\,$. Then, we'd need to solve: $$ \cos D = 0.5 $$ So the question is: Of course, there are infinitely many: any angle whose terminal point has an $x$value of $\,0.5\,$. (Look right and/or below to see a bunch of them.) $$ \begin{alignat}{4} \cos 60^\circ &= 0.5& \qquad \cos(60^\circ) &= 0.5& \qquad \cos(300^\circ) &= 0.5& \qquad \cos 300^\circ &= 0.5\cr \cos 420^\circ &= 0.5& \qquad \cos(420^\circ) &= 0.5& \qquad \cos(660^\circ) &= 0.5& \qquad \cos(1740^\circ) &= 0.5 \end{alignat} $$ The arccosine function can be abbreviated as ‘$\,\arccos\,$’, but is still pronounced ‘arccosign’. Arccosine is a function: each input has exactly one output. When we ask for $\,\text{arccos}(0.5)\,$, we can't be given all the angles with a cosine of $\,0.5\,$! We need $\,\text{arccos}(0.5)\,$ to represent exactly one angle. So, which one? Answer: the one between $\,0^\circ\,$ and $\,180^\circ\,$! That is, by definition: $$ \text{arccos}(x) = \text{the angle between $\,0^\circ\,$ and $\,180^\circ\,$ whose cosine is $x$} $$ So, what exactly is $\,\text{arccos}(x)\,$? Answer: It is the angle between $\,0^\circ\,$ and $\,180^\circ\,$ that has a cosine of $\,x\,$. So: $$\text{arccos}(0.5) \ =\ \text{the angle between $\,0^\circ\,$ and $\,180^\circ\,$ whose cosine is $0.5$} \ =\ 60^\circ$$ 
$\theta = 60^\circ\,$, $\cos\theta = 0.5$ $\theta = 300^\circ\,$, $\cos\theta = 0.5$ 
$\theta = 60^\circ\,$, $\cos\theta = 0.5$ $\theta = 300^\circ\,$, $\cos\theta = 0.5$ 
At this point you've been introduced
to both the arcsine and the arccosine.
The arcsine ‘undoes’ the sine.
The arccosine ‘undoes’ the cosine.
Let's compare them:


A triangle has sides of lengths $\,3\,$, $\,5\,$ and $\,7\,$. Solve the triangle. Solution:






On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
