\chapter{Analytic trigonometry}
\section{Law of cosines}
In the previous unit we used the law of sines to solve triangles when two angles and one side are
given (ASA case) or when two sides and one of the opposite sides are given (SSA case).
There are two more possible cases: if we are given two sides and the angle between them (SAS case):
|
\(\img{U5_7F3.png}{}{14em}{}\) |
|
SAS case |
and if we are given the three sides (SSS case):
|
\(\img{U5_7F4.png}{}{14em}{}\) |
|
SSS case |
We will solve these triangles using the law of cosines. Recall the Pythagorean theorem for right triangles:
\[
\begin{array}{cc}
\img{U5_7F1.png}{-5em}{14em}{}&
c^2=a^2+b^2
\end{array}
\]
The Law of Cosines is a generalization of the Pythagorean theorem that works for any triangle:
\[
\begin{array}{cc}
\img{U5_7F2.png}{-3em}{14em}{}& \begin{array}{c}
c^2=a^2+b^2-2ab\cos C\\
a^2=b^2+c^2-2bc \cos A\\
b^2=c^2+a^2 -2ac \cos B
\end{array}
\end{array}
\]
There is an important difference between the sine and the cosine functions: the inverse sine
cannot directly give us the obtuse angles (in Q2), because its range is \([-\pi/2,\pi/2]\). But
the inverse cosine will give us obtuse angles solutions, because its range is \([0,\pi]\). When
solving the SAS and SSS cases, we will need to bear this in mind in our choice of sides or angles
to be solved first.
\subsection{The SAS case}
When we know two sides of a triangle and the angle between them (that must be less than
\(180^\circ\) of course), there will always be a unique solution. This is called the SAS case.
In this case, there is no choice for the first step: we can only use the law of cosines
for the given angle.
\begin{example}
We want to solve the triangle with \(A=36^\circ\), \(b=6\), \(c=11\).
\[\img{U5_7F5.png}{}{14em}{}\]
We use the law of cosines for angle \(A\):
\[\begin{array}{rcl}
a^2&=& b^2+c^2-2bc\cos A \\[1ex]
a^2&=& 6^2 +11^2 -2(6)(11)\cos(36^\circ)\\[1ex]
a^2&=& 36+121-132 \cos(36^\circ)\\[1ex]
a^2&\approx& 50.210\\[1ex]
a&\approx& \sqrt{50.21}\\[1ex]
a&\approx& 7.086
\end{array}
\]
To find the other two angles, we use the law of sines. Whenever you use the law of sines and have
a choice between two angles, choose the angle opposite the
\textbf{smaller} side. This is to avoid that the angle we are trying to find may be obtuse, in which
case the inverse sine function will give us an answer in Q1, that would not be the right answer.
So we solve for angle \(B\) opposite side \(b\):
\[\begin{array}{rcl}
\displaystyle\frac{\sin B}{b}&=& \displaystyle\frac{\sin A}{a}\\[2ex]
\displaystyle \frac{\sin B}{6} &=& \displaystyle \frac{\sin 36^\circ }{7.086} \\[2ex]
\sin B &=& \displaystyle 6\cdot \frac{\sin 36^\circ }{7.086} \\[2ex]
\sin B &\approx& 0.498\\[2ex]
B&\approx& \sin^{-1}(0.498)\\[2ex]
B&\approx& 29.848^\circ
\end{array}
\]
The third angle is then found by subtracting the two angles from \(180^\circ\):
\[C=180^\circ - 36^\circ -29.848^\circ=114.152^\circ\]
\end{example}
\subsection{The SSS case}
When we are given three positive numbers \(a,b,c\), there will be a unique triangle
with those sides as long as no number is larger than the sum of the other two. That's because
in any triangle, the sum of two sides must be larger than the third side. So for example
it's impossible to form a triangle with sides \(1,2,4\), because \(4\) is larger than \(1+2\).
The starting point is to use the law of cosines to find one of the three unknown angles.
Then we use the law of sines to find a second angle. Since
the law of cosines will give us directly obtuse angles (in Q2) if there are any, make
sure to always start with the angle opposite the \textbf{largest} side when using the law of
cosines. This way, if there is an obtuse angle, we find it right away, and then the remaining angles
must be acute, so there will be no danger of mistakes when using the law of sines to find them.
\begin{example}
We solve the triangle in the picture:
\[\img{U5_7F6.png}{}{14em}{}\]
We use the law of cosines for the angle opposite the longest side:
\[\begin{array}{rcl}
23^2&=& 18^2+10^2-2(18)(10)\cos C \\[2ex]
529&=& 324 +100 -360\cos C\\[2ex]
105&=& -360 \cos C\\[2ex]
\cos C&=& \displaystyle -\frac{105}{360}\\[2ex]
\cos C&\approx& -0.2917\\[2ex]
C&\approx& \cos^{-1}(-0.2917)\\[2ex]
C&\approx& 106.96^\circ
\end{array}
\]
Use the law of sines:
\[\begin{array}{rcl}
\displaystyle\frac{\sin B}{b}&=& \displaystyle\frac{\sin C}{c}\\[2ex]
\displaystyle \frac{\sin B}{18} &=& \displaystyle \frac{\sin 106.96^\circ }{23} \\[2ex]
\sin B &=& \displaystyle 18\cdot \frac{\sin 106.96^\circ }{23} \\[2ex]
\sin B &\approx& 0.749\\[2ex]
B&\approx& \sin^{-1}(0.749)\\[2ex]
B&\approx& 48.467^\circ
\end{array}
\]
Then the third angle is found by subtracting from \(180^\circ\):
\[A=180^\circ - 106.96^\circ -48.467^\circ=24.573^\circ\]
Note that in this example angle \(C\) is obtuse. This means that if we had not started with
it and we had instead found either \(A\) or \(B\) first, then using the law of sines and the
\(\sin^{-1}\) function to find \(C\) would have incorrectly given an answer in Q1 for angle \(C\).
\end{example}
\textbf{Law of cosines}
|
\(c^2=a^2+b^2-2ab\cos C\) |
| \(a^2=b^2+c^2-2bc \cos A\) |
| \(b^2=c^2+a^2 -2ac \cos B\) |
\textbf{SAS case}
- Use the law of cosines to find the side opposite the given angle.
- Use the law of sines for the \textbf{smaller} of the other two angles.
- Find the third angle by subtracting from \(180^\circ\).
|
\textbf{SSS case}
- Use the law of cosines to find the angle opposite the \textbf{largest} side.
- Use the law of sines to find one of the other two angles.
- Find the third angle by subtracting from \(180^\circ\).
|
Problems
\problem
Solve the triangle with \(A=39^\circ\), \(b=28\), \(c=7\).
\begin{sol}
This is a SAS case. We solve for \(a\) using the law of cosines:
\[\begin{array}{rcl}
a^2&=& b^2+7c^2-2bc\cos A \\[2ex]
a^2&=& 28^2 +7^2 -2(28)(7)\cos(39^\circ)\\[2ex]
a^2&=& 784+49-392 \cos(39^\circ)\\[2ex]
a^2&\approx& 528.359\\[2ex]
a&\approx& \sqrt{50.21}\\[2ex]
a&\approx& 22.986
\end{array}
\]
Now we use the law of sines, making sure to choose the angle opposite the \textbf{smaller} side:
\[\begin{array}{rcl}
\displaystyle\frac{\sin C}{c}&=& \displaystyle\frac{\sin A}{a}\\[2ex]
\displaystyle \frac{\sin C}{7} &=& \displaystyle \frac{\sin 39^\circ }{22.986} \\[2ex]
\sin C &=& \displaystyle 7\cdot \frac{\sin 39^\circ }{22.986} \\[2ex]
\sin C &\approx& 0.192\\[2ex]
C&\approx& \sin^{-1}(0.192)\\[2ex]
C&\approx& 11.049^\circ
\end{array}
\]
and then we find the third angle by subtracting from \(180^\circ\):
\[B=180^\circ-39^\circ-11.049^\circ=129.951^\circ.\]
So the solution is
\[\fbox{\(a\approx 22.986\)} \hspace{2ex} \fbox{\(C\approx 11.049^\circ\)} \hspace{2ex}
\fbox{\(B\approx 129.951^\circ\)}\]
\end{sol}
\mproblem
Solve the triangle with \(B=46^\circ\), \(a=19\), \(c=11\).
\problem
Solve the triangle with \(a=13\), \(b=29\), \(c=37\).
\begin{sol}
This is a SSS case. We start using the law of cosines for the \textbf{largest} side:
\[\begin{array}{rcl}
37^2&=& 29^2+13^2-2(29)(13)\cos C \\[2ex]
1369&=& 841 +169 -754\cos C\\[2ex]
359&=& -754 \cos C\\[2ex]
\cos C&=& \displaystyle -\frac{359}{754}\\[2ex]
\cos C&\approx& -0.476\\[2ex]
C&\approx& \cos^{-1}(-0.476)\\[2ex]
C&\approx& 118.433^\circ
\end{array}
\]
Now use the law of sines:
\[\begin{array}{rcl}
\displaystyle\frac{\sin B}{b}&=& \displaystyle\frac{\sin C}{c}\\[2ex]
\displaystyle \frac{\sin B}{29} &=& \displaystyle \frac{\sin 118.433^\circ }{37} \\[2ex]
\sin B &=& \displaystyle 29\cdot \frac{\sin 118.433^\circ }{37} \\[2ex]
\sin B &\approx& 0.689\\[2ex]
B&\approx& \sin^{-1}(0.689)\\[2ex]
B&\approx& 43.570^\circ
\end{array}
\]
The third angle is
\[A=180^\circ-118.433^\circ-43.570^\circ=17.997^\circ.\]
So the solution is
\[\fbox{\(C\approx 118.433^\circ\)} \hspace{2ex} \fbox{\(B\approx 43.570^\circ\)} \hspace{2ex}
\fbox{\(A\approx 17.997^\circ\)}\]
\end{sol}
\mproblem
Solve the triangle with \(a=27\), \(b=17\), \(c=20\).