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 (side-side-side) or SAS (side-angle-side) 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: $$ \cssId{s12}{d^2 = e^2 + f^2 - 2ef\cos D} $$ Solving for $\cos D\,$ gives $$ \cssId{s14}{\cos D = \frac{d^2 - e^2 - f^2}{-2ef}} $$ The right-hand side is just a constant, since $\,d\,,$ $\,e\,$ and $\,f\,$ are all known. For the moment, suppose that this right-hand side evaluates to (say) $\,0.5\,.$ Then, we'd need to solve: $$ \cssId{s18}{\cos D = 0.5} $$ 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.) $$\cssId{s24}{ \begin{alignat}{4} \cos 60^\circ &= 0.5& \qquad \cos(-60^\circ) &= 0.5& \qquad \cos(-300^\circ) &= 0.5& \qquad \cos 300^\circ &= 0.5\cr \cos 420^\circ &= 0.5& \qquad \cos(-420^\circ) &= 0.5& \qquad \cos(660^\circ) &= 0.5& \qquad \cos(-1740^\circ) &= 0.5 \end{alignat}} $$ The arccosine function can be abbreviated as ‘$\,\arccos\,$’, but is still pronounced ‘arc-co-sign’. 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: $$ \cssId{s34}{\text{arccos}(x) = \text{the angle between $\,0^\circ\,$ and $\,180^\circ\,$ whose cosine is $x$}} $$ 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: $$ \cssId{s39}{\text{arccos}(0.5) \ =\ \text{the angle between $\,0^\circ\,$ and $\,180^\circ\,$ whose cosine is $0.5$} \ =\ 60^\circ} $$ |
![]() $\theta = 60^\circ\,,$ $\cos\theta = 0.5$ ![]() $\theta = -300^\circ\,,$ $\cos\theta = 0.5$ |
![]() $\theta = -60^\circ\,,$ $\cos\theta = 0.5$ ![]() $\theta = 300^\circ\,,$ $\cos\theta = 0.5$ |
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:
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A triangle has sides of lengths $\,3\,,$ $\,5\,$ and $\,7\,.$ Solve the triangle. Solution:
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On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. |
PROBLEM TYPES:
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