\chapter{Analytic trigonometry}
\section{Law of sines}
So far we have discussed \textbf{right} triangles (where one of the angles is \(90^\circ\)).
In this unit and the next, we discuss general triangles.
We will draw a triangle by using capital letters \(A,B,C\)
for the angles, and lower case letters \(a,b,c\) for the sides, and in such a way that
a side opposite an angle is marked with the same letter as the angle (see the picture):
\[\img{U5_6F1.png}{}{20em}{}\]
The Law of Sines is a set of three equations that can be used to solve for the sides or angles
of a triangle. The three equations are:
\[\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\]
Note that this is really three equations:
\[\frac{a}{\sin A}=\frac{b}{\sin B}, \hspace{5ex} \frac{b}{\sin B}=\frac{c}{\sin C}, \hspace{5ex}
\frac{a}{\sin A}=\frac{c}{\sin C}.\]
We use this set when we are solving for the side of a triangle. If we need to solve
for an angle, it is more convenient to use the reciprocal of each fraction:
\[\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}.\]
There are two different types of problems we can solve with the Law of Sines, labeled AAS and SSA.
S stands for side, and A for angle. So AAS means that we are given two angles and one side,
and SSA means that we are given two sides and one of the opposite angles
(so it will not be the angle between the two given sides).
\subsection{The AAS case}
If we know two angles and one side, then (as long a the sum of the two given angles is less than
\(180^\circ\)), there is always exactly one solution.
\begin{example}
Suppose we are given \(a=7\), \(C=31^\circ\), \(A=70^\circ\). We draw a picture by showing as solid
lines the given sides or angles, and by dashed lines the unknowns.
\[\img{U5_6F2.png}{}{20em}{}\]
Since we already know two angles, it is easy to find the third, by using the fact that
\(A+B+C=180^\circ\). So we subtract the given angles from \(180^\circ\) and
we find \(B=180-31-70=79^\circ\).
Now we use one of the three equations that has only one remaining unknown:
\[\begin{array}{rcl}
\displaystyle \frac{b}{\sin B}&=& \displaystyle\frac{a}{\sin A}\\[2ex]
\displaystyle \frac{b}{\sin 79^\circ}&=& \displaystyle \frac{7}{\sin 70^\circ}
\end{array}
\]
To find \(b\), multiply by \(\sin 79^\circ\):
\[\begin{array}{rcl}
b&=& \displaystyle \sin 79^\circ \cdot \frac{7}{\sin 70^\circ}\\[1ex]
&\approx& 7.312
\end{array}\]
To find \(c\), use
\[\begin{array}{rcl}
\displaystyle \frac{c}{\sin C}&=& \displaystyle\frac{a}{\sin A}\\[2ex]
\displaystyle\frac{c}{\sin 31^\circ}&=& \displaystyle\frac{7}{\sin 70^\circ}
\end{array}
\]
Multiply by \(\sin 31^\circ\):
\[\begin{array}{rcl}
c&=& \displaystyle \sin 31^\circ \cdot \frac{7}{\sin 70^\circ}\\[1ex]
&\approx& 3.837.
\end{array}\]
We have now solved the triangle, as shown below:
\[\img{U5_6F9.png}{}{20em}{}\]
\end{example}
\subsection{The SSA case}
If we are given two sides and an angle that is not between the two sides, the problem
is more complicated, because there could be one solution, or two solutions, or no solutions at all.
To understand why, look at the next picture, that shows the case of a single solution:
\[\img{U5_6F3.png}{}{20em}{}\]
Remember that the solid lines represent the quantities that are given. In the picture
above there is a solution for the triangle. But if the given side \(c\) were much shorter,
as in the following picture:
\[\img{U5_6F4.png}{}{10em}{}\]
then
it would be impossible to form a triangle, no matter how we change the dashed angles and sides,
because side \(c\) is just too short and cannot reach side \(b\) to form a triangle.
On the other hand, if side \(c\) is a bit longer than the height
\(h\) of the triangle, but still shorter than \(a\),
then we could form two different triangles with the given data,
by placing \(c\) either to the right or to the left of the line through the
top vertex and perpendicular to \(b\):
\[\img{U5_6F5.png}{}{10em}{}\hspace{10ex} \img{U5_6F6.png}{}{10em}{}\]
You can watch an animation of the problem on
this website.
\begin{example}
Suppose we are given \(a=12, c=22, C=42^\circ\).
\[\img{U5_6F7.png}{}{20em}{}\]
Using the ratios for \(a\) and \(c\), the only unknown will be the angle \(A\), so we place
the sine functions in the numerator:
\[\begin{array}{rcl}
\displaystyle\frac{\sin A}{a}&=& \displaystyle\frac{\sin C}{c}\\[1ex]
\displaystyle \frac{\sin A}{12} &=& \displaystyle \frac{\sin 42^\circ }{22} \\[1ex]
\sin A &=& \displaystyle 12\cdot \frac{\sin 42^\circ }{22} \\[1ex]
\sin A &\approx& 0.37
\end{array}
\]
Using the \(\sin^{-1}\) key, we find
\(A\approx 21.40^\circ\).
But remember that the \(\sin^{-1}\) key will only give us angles in \([-90^\circ,90^\circ]\).
There is also a solution to the equation \(\sin A = 0.365\) in Q2, with reference angle
\(21.40^\circ\):
\[\img{U5_6F8.png}{}{15em}{}\]
The second solution for \(A\) is \(180^\circ - 21.40^\circ=158.6^\circ\). But this second solution is not
a valid value for angle \(A\) of
the original triangle, because if we add it to the given angle \(C\) we find
\(158.6^\circ+42^\circ= 200.6^\circ\), that is already larger than \(180^\circ\), so it is impossible
to form a triangle.
So we conclude that the only valid solution for \(A\) is \(A=21.40^\circ\).
Then we find \(B=180^\circ -A-C=180^\circ -21.40^\circ -42^\circ=116.6^\circ\). To find \(b\),
we use the Law of Sines again:
\[
\begin{array}{rcl}
\displaystyle \frac{b}{\sin B}&=& \displaystyle\frac{c}{\sin C}\\[2ex]
\displaystyle \frac{b}{\sin (116.6^\circ)}&=& \displaystyle \frac{22}{\sin (42^\circ)}\\[2ex]
b&=& \displaystyle \sin (116.6^\circ)\cdot \frac{22}{\sin(42^\circ)}\\[2ex]
b&\approx & 29.4
\end{array}
\]
\[\img{U5_6F10.png}{}{16em}{}\]
\end{example}
\begin{example}
We are given the SSA case \(a=25\), \(c=15\), \(C=85^\circ\). We find:
\[\begin{array}{rcl}
\displaystyle\frac{\sin A}{a}&=& \displaystyle\frac{\sin C}{c}\\[1ex]
\displaystyle \frac{\sin A}{25} &=& \displaystyle \frac{\sin 85^\circ }{15} \\[1ex]
\sin A &=& \displaystyle 25\cdot \frac{\sin 85^\circ }{15} \\[1ex]
\sin A &=& 1.66\ldots
\end{array}
\]
But this last equation is impossible, because \(\sin A\) must be in the interval \([-1,1]\).
So this problem has no solution.
\end{example}
\begin{example}
We want to find a triangle with \(c=31\), \(b=12\), \(B=20^\circ\). Using the ratio for \(b\) and
\(c\), we find:
\[\begin{array}{rcl}
\displaystyle\frac{\sin C}{c}&=& \displaystyle\frac{\sin B}{b}\\[2ex]
\displaystyle \frac{\sin C}{31} &=& \displaystyle \frac{\sin 20^\circ }{12} \\[2ex]
\sin C &=& \displaystyle 31\cdot \frac{\sin 20^\circ }{12} \\[2ex]
\sin C &\approx& 0.88
\end{array}
\]
Using the \(\sin^{-1}\) key, this gives us
\(C\approx 62.07^\circ\).
But there is a second solution in Q2, given by \(180^\circ -62.07=117.93^\circ\). This time
the second solution is valid, because \(C+B=117.93^\circ+20^\circ=137.93^\circ\), and there is
room for the third angle \(A\).
So we are going to have two different triangles, with different
values for \(C\), \(A\) and \(a\). We label \(C_1=62.07^\circ\) the first value for \(C\)
and \(C_2=117.93^\circ\) the second. Then we have corresponding \(A\) values
| \(A_1\) | \(=\) | \(180^\circ-B-C_1\) |
| \(=\) | \(180^\circ-20^\circ-62.07^\circ\) |
| \(=\) | \(97.93^\circ\) |
| and |
| \(A_2\) | \(=\) | \(180^\circ - B-C_2\) |
| \(=\) | \(180^\circ -20^\circ-117.93^\circ\) |
| \(=\) | \(42.07^\circ\). |
|
To find \(a\), use the ratio for \(a\) and \(b\) in each case:
| \(\displaystyle \frac{a_1}{\sin A_1}=\displaystyle\frac{b}{\sin B}\) |
| \(\displaystyle \frac{a_1}{\sin (97.93^\circ)}= \frac{12}{\sin (20^\circ)}\) |
| \(a_1 =\displaystyle \sin (97.93^\circ)\cdot \frac{12}{\sin(20^\circ)}\) |
| \(a_1\approx 34.75\) |
| and |
| \(\displaystyle \frac{a_2}{\sin A_2}==\displaystyle\frac{b}{\sin B}\) |
| \(\displaystyle \frac{a_2}{\sin (42.07^\circ)}=\displaystyle \frac{12}{\sin (20^\circ)}\) |
| \(a_2= \displaystyle \sin (42.07^\circ)\cdot \frac{12}{\sin(20^\circ)}\) |
| \(a_2\approx 23.51\) |
|
The picture below shows the two different triangles.
\[\img{U5_6F11.png}{}{20em}{} \hspace{2ex} \img{U5_6F12.png}{}{18em}{}\]
\end{example}
\textbf{Law of sines}
To solve for sides: \(\hspace{3ex} \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\)
|
| To solve for angles: \( \hspace{3ex} \dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}\)
|
\textbf{AAS case}
| One solution if the sum of the two given angles is less than \(180^\circ\), otherwise no solutions. |
\textbf{SSA case}
| No solution: | \(\img{U5_6F4.png}{-5em}{8em}{}\) |
| One solution: | \(\img{U5_6F3.png}{-5em}{14em}{}\)
|
| Two solutions: | \(\img{U5_6F5.png}{-5em}{8em}{}\hspace{10ex}
\img{U5_6F6.png}{-5em}{8em}{}\) |
Problems
\problem
Solve the triangle in the picture, and approximate the answer to two decimals:
\[\img{U5_6F13.png}{}{25em}{}\]
\begin{sol}
This is the AAS case. First we find the third angle:
\[
A= 180^\circ - 20^\circ - 40^\circ\\
\fbox{\(A= 120^\circ\)}
\]
Then we use the ratios for \(a\) and \(b\):
\[\begin{array}{rcl}
\displaystyle \frac{a}{\sin A}&=& \displaystyle\frac{b}{\sin B}\\[2ex]
\displaystyle \frac{18}{\sin 120^\circ}&=& \displaystyle \frac{b}{\sin 20^\circ}\\[2ex]
b&=& \displaystyle \sin(20^\circ) \cdot \frac{18}{\sin(120^\circ)}
\end{array}
\\
\fbox{\(b\approx 7.11\)}
\]
To find \(c\), we use the ratio for \(a\) and \(c\):
\[\begin{array}{rcl}
\displaystyle \frac{a}{\sin A}&=& \displaystyle\frac{c}{\sin C}\\[2ex]
\displaystyle \frac{18}{\sin 120^\circ}&=& \displaystyle \frac{c}{\sin 40^\circ}\\[2ex]
c&=& \displaystyle \sin(40^\circ) \cdot \frac{18}{\sin(120^\circ)}
\end{array}\\
\fbox{\(c\approx 13.36\)}
\]
\end{sol}
\mproblem
Solve the triangle in the picture, and approximate the answer to two decimals:
\[\img{U5_6F14.png}{}{20em}{}\]
\problem
Solve the triangle \(ABC\) given that \(c=31\), \(b=29\) and \(B=64^\circ\).
\begin{sol}
This is the \(SSA\) case. We use the ratio for \(b\) and \(c\):
\[\begin{array}{rcl}
\displaystyle\frac{\sin C}{c}&=& \displaystyle\frac{\sin B}{b}\\[2ex]
\displaystyle \frac{\sin C}{31} &=& \displaystyle \frac{\sin 64^\circ }{29} \\[2ex]
\sin C &=& \displaystyle 31\cdot \frac{\sin 64^\circ }{29} \\[2ex]
\sin C &\approx& 0.96
\end{array}
\]
Using the \(\sin^{-1}\) key, we find \(C=73.90^\circ\). The corresponding angle in Q2 is
\(180^\circ-73.90^\circ=106.10^\circ\). Since the given angle is \(64^\circ\) and
\(106.10+64\) is less than \(180\), there is a second solution. So we have
\[\fbox{\(C_1=73.90^\circ\)}, \hspace{3ex} \fbox{\(C_2=106.10^\circ\).}\]
The corresponding values for \(A\) are
\[A_1=180^\circ -64^\circ - 73.90^\circ=42.1^\circ, \\
A_2=180^\circ - 64^\circ - 106.10^\circ = 9.9^\circ.\]
Now we can find \(a_1\) and \(a_2\):
\[\begin{array}{rcl}
\displaystyle \frac{a_1}{\sin A_1}&=& \displaystyle\frac{b}{\sin B}\\[2ex]
\displaystyle \frac{a_1}{\sin 42.1^\circ}&=& \displaystyle \frac{29}{\sin 64^\circ}\\[2ex]
a_1&=& \displaystyle \sin(42.1^\circ) \cdot \frac{29}{\sin(64^\circ)}
\end{array}\\
\fbox{\(a_1\approx 21.63\)}
\]
\[\begin{array}{rcl}
\displaystyle \frac{a_2}{\sin A_2}&=& \displaystyle\frac{b}{\sin B}\\[2ex]
\displaystyle \frac{a_2}{\sin 9.9^\circ}&=& \displaystyle \frac{29}{\sin 64^\circ}\\[2ex]
a_2&=& \displaystyle \sin(9.9^\circ) \cdot \frac{29}{\sin(64^\circ)}
\end{array}\\
\fbox{\(a_2\approx 5.55\)}
\]
\end{sol}
\mproblem
Solve the triangle \(ABC\) given that \(c=19\), \(b=15\) and \(B=25^\circ\).
\problem
Solve the triangle \(ABC\) given that \(a=20\), \(c=5\) and \(C=35^\circ\).
\begin{sol}
We set up the equation to find angle \(A\):
\[\begin{array}{rcl}
\displaystyle\frac{\sin A}{a}&=& \displaystyle\frac{\sin C}{c}\\[2ex]
\displaystyle \frac{\sin A}{20} &=& \displaystyle \frac{\sin 35^\circ }{5} \\[2ex]
\sin A &=& \displaystyle 20\cdot \frac{\sin 35^\circ }{5} \\[2ex]
\sin A &=& 2.29\ldots
\end{array}
\]
This equation is impossible, because \(\sin A\) must be in the interval \([-1,1]\). So this
problem has no solution.
\end{sol}
\mproblem
Solve the triangle \(ABC\) given that \(b=18\), \(a=4\) and \(A=31^\circ\).
\problem
Solve the triangle \(ABC\) given that \(a=20\), \(c=28\) and \(C=62^\circ\).
\begin{sol}
We solve for angle \(A\):
\[\begin{array}{rcl}
\displaystyle\frac{\sin A}{a}&=& \displaystyle\frac{\sin C}{c}\\[2ex]
\displaystyle \frac{\sin A}{20} &=& \displaystyle \frac{\sin 62^\circ }{28} \\[2ex]
\sin A &=& \displaystyle 20\cdot \frac{\sin 62^\circ }{28} \\[2ex]
\sin A &\approx& 0.631
\end{array}
\]
Using the \(\sin^{-1}\) key, we find
\[\fbox{\(A=39.10^\circ\).}\]
The solution in Q2 is
\(180^\circ-39.10=140.90^\circ\). But the given angle is \(62^\circ\), and \(62+140.90\)
is more than \(180\), so the angle in Q2 does not give us a second triangle.
We then find \(180^\circ-39.10^\circ - 62^\circ = 78.9^\circ\), so
\[\fbox{\(B=78.9^\circ\)}\]
and
\[\begin{array}{rcl}
\displaystyle \frac{b}{\sin B}&=& \displaystyle\frac{c}{\sin C}\\[2ex]
\displaystyle \frac{b}{\sin 78.9^\circ}&=& \displaystyle \frac{28}{\sin 62^\circ}\\[2ex]
b&=& \displaystyle \sin(78.9^\circ) \cdot \frac{28}{\sin(62^\circ)}
\end{array}\\
\fbox{\(b\approx 31.12\)}
\]
\end{sol}
\mproblem
Solve the triangle \(ABC\) given that \(b=13\), \(a=28\) and \(A=55^\circ\).