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\noindent
{\large\bf `Solving' Triangles:}

A triangle has three sides and three angles, and this is 
six pieces of information. Typically if you know three
of these pieces of information then you can calculate
the other three using properties of the trig functions.
This has practical applications.
Some distances are hard to calculate, such as
the distance between two mountain peaks or two stars.
If one is lucky then it may be possible to 
realize that hard-to-measure
distance as one side of a triangle
where at least one of
the other two sides can be measured more easily.
Then, after measuring an angle or two, one can
calculate the hard-to-measure distance.
Our book calls this type of calculation 
`solving a triangle'.
The mathematical tools needed
for these calculations are 
\begin{itemize}
\item the Law of Sines, and
\item the Law of Cosines.
\end{itemize}

\bigskip

\noindent
{\large\bf The Law of Sines:}

\bigskip

The Law of Sines is a byproduct of the area formula
for a triangle.

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\put(36,18){$h$}
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In this picture the measure of the base is $b$ while the height
is $h = c\sin(A)$. Therefore the area is $\frac{1}{2}{\rm
(base)(height)} = \frac{1}{2}bc\sin(A)$. This is the area
calculation from the `viewpoint' of angle $A$. Measuring
the area from the viewpoint of each of the angles we get that
\[Area = \frac{1}{2}bc\sin(A) = 
\frac{1}{2}ac\sin(B) = 
\frac{1}{2}ab\sin(C).\]
Dividing through by $\frac{1}{2}abc$ yields the 
Law of Sines:
\[\frac{\sin(A)}{a} = 
\frac{\sin(B)}{b} = 
\frac{\sin(C)}{c}.\]

\noindent
{\bf Exercise.} (Calculator needed.)
Suppose that $A = \pi/4, C = \pi/3$ and $b = 10$ feet.
Find $a, c, B, h$ and the area of the triangle.
Use the fact that for any triangle
the sum of the angles $A + B + C = \pi$.


\break

\noindent
{\large\bf The Law of Cosines:}

\bigskip

The Law of Cosines is a generalization of the Pythagorean Theorem.

\[\begin{array}{rl}
a^2 &= b^2 + c^2 - 2bc\cos(A)\\
b^2 &= a^2 + c^2 - 2ac\cos(B)\\
c^2 &= a^2 + b^2 - 2ab\cos(C).
\end{array}\]
In fact, it is proved by applying the Pythagorean Theorem
to the two small triangles in the preceding picture:

\[\begin{array}{rl}
c^2 &= h^2 + (c\cos(A))^2\\
a^2 &= h^2 + (b-c\cos(A))^2
\end{array}\]
Subtracting on both sides we get 
$c^2-a^2 = (c\cos(A))^2 - (b-c\cos(A))^2$
which simplifies to 
$a^2 = b^2 + c^2 - 2bc\cos(A)$, one form of the 
Law of Cosines.

\bigskip

\noindent
{\bf Exercise.} (Calculator needed.)
Suppose that $A = \pi/4, b = 10$ feet and
$c = 6$ feet.
Find $a, B, C, h$ and the area of the triangle.

\vspace{2in}

\noindent
The Law of Sines and the Law of Cosines can be used
together to calculate the distance between two 
mountain peaks or two stars. (Draw a picture.)


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