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 the Law of Sines requires knowing an angle and its opposite side.
Therefore, we need a different tool to begin solving a
SSS (sidesideside) or SAS (sideangleside) configuration.
The Law of Cosines (which can be viewed as a generalization of the Pythagorean Theorem) comes to the rescue!
In this section, capital letters are used to denote angles/vertices. Corresponding lowercase letters are used to denote lengths of opposite sides. Thus:
To significantly cut down on words, I say things like:


Consider the triangle at right. Since $\,D\,$ is a right angle, the wellknown Pythagorean Theorem tells us: $$\cssId{s19}{d^2 = e^2 + f^2}$$ Such a beautiful, simple, relationship between the lengths of the sides: $$ \cssId{s21}{\overbrace{d^2\strut }^{\text{the square of the longest side}}} \cssId{s22}{\overbrace{=\strut }^{\text{is}}} \cssId{s23}{\overbrace{e^2 + f^2\strut }^{\text{the sum of the squares of the shorter sides}}} $$ 

But, what if $\,D\,$ isn't a right angle?
Let's investigate. Keep $\,e\,$ and $\,f\,$ the same, but change angle $\,D\,$:
the Pythagorean Theorem to account for the change in the length of $\,d\,$! 
If $\,D\,$ becomes an obtuse angle, its opposite side gets longer: $$\cssId{s35}{d_{\text{new}} > d}$$ 
If $\,D\,$ becomes an acute angle, its opposite side gets shorter: $$\cssId{s37}{d_{\text{new}} < d}$$ 
Consider an arbitrary triangle (see right), where:


For angles $\,D\,$ strictly between $\,90^\circ\,$ and $\,180^\circ\,$, we have $\,\cos D < 0\,$. Since $\,e\,$ and $\,f\,$ (side lengths) are strictly positive, this means that $$ \cssId{s85}{\text{the adjustment}} \cssId{s86}{\quad=\quad} \cssId{s87}{\overbrace{2\strut}^{()}} \ \cssId{s88}{\overbrace{ef\strut}^{(+)}} \ \cssId{s89}{\overbrace{\cos D\strut}^{()}} \quad \cssId{s90}{> \quad 0} $$ Since the adjustment is positive, $\,d\,$ is longer than it would be, if it were opposite a $\,90^\circ\,$ angle. Take another look at the image from the motivation (at right)! 

For angles $\,D\,$ strictly between $\,0^\circ\,$ and $\,90^\circ\,$, we have $\,\cos D > 0\,$. Since $\,e\,$ and $\,f\,$ (side lengths) are strictly positive, this means that $$ \cssId{s97}{\text{the adjustment}} \cssId{s98}{\quad=\quad} \cssId{s99}{\overbrace{2\strut}^{()}}\ \cssId{s100}{\overbrace{ef\strut}^{(+)}} \cssId{s101}{\ \overbrace{\cos D\strut}^{(+)}} \cssId{s102}{\quad < \quad 0}$$ Since the adjustment is negative, $\,d\,$ is shorter than it would be, if it were opposite a $\,90^\circ\,$ angle. Take another look at the image from the motivation (at right)! 

The proof of the Law of Cosines is a smidgeon harder than the proof of the Law of Sines,
but it's not too bad for such a powerful result.
Here's a renaming of rectangular coordinates that is needed in the proof:




Consider a triangle with side lengths $\,3\,$ and $\,5\,$, and an included $\,57^\circ\,$ angle.
Prove that this information uniquely defines a triangle, and then solve the triangle.
If using a calculator, store the full accuracy of computed sides/angles.
Then, use these (most accurate) stored values in intermediate calculations.
Summarize the values of sides and angles in a table.
In the table, round values to one decimal place.
In the solution below, the default accuracy provided by WolframAlpha is reported and used for intermediate calculations.
By
SAS, the triangle is unique.
The notation at right is used in the solution.


If you (like me) want everything to be ‘just perfect’ in your mathematical world,
then you may have been bothered by the usage of ‘$\,=\,$’ and ‘$\,\approx\,$’ in the previous example.
If so, click here!
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
