2.1 The Law of Sines

Theorem 2.1

Law of Sines: If a triangle has sides of lengths $a$, $b$, and $c$ opposite the angles $A$, $B$, and $C$, respectively, then

\begin{equation} \frac{a}{\sin\;A} ~=~ \frac{b}{\sin\;B} ~=~ \frac{c}{\sin\;C} ~. \tag{2.1}\end{equation}

Note that by taking reciprocals, equation (2.1) can be written as

\begin{equation} \frac{\sin\;A}{a} ~=~ \frac{\sin\;B}{b} ~=~ \frac{\sin\;C}{c} ~, \tag{2.2}\end{equation}

and it can also be written as a collection of three equations:

\begin{equation} \frac{a}{b} ~=~ \frac{\sin\;A}{\sin\;B} ~~,\quad \frac{a}{c} ~=~ \frac{\sin\;A}{\sin\;C} ~~,\quad \frac{b}{c} ~=~ \frac{\sin\;B}{\sin\;C} \tag{2.3}\end{equation}

Another way of stating the Law of Sines is: The sides of a triangle are proportional to the sines of their opposite angles.

To prove the Law of Sines, let $\triangle\,ABC$ be an oblique triangle. Then $\triangle\,ABC$ can be acute, as in Figure 2.1.1(a), or it can be obtuse, as in Figure 2.1.1(b). In each case, draw the altitude^[1] from the vertex at $C$ to the side $\overline{AB}$. In Figure 2.1.1(a) the altitude lies inside the triangle, while in Figure 2.1.1(b) the altitude lies outside the triangle.

tikzpicture[every node/.style=font=] [fillcolor] (0,0) -- (1.8,2) -- (3,0) -- cycle; [dashed] (1.8,2) -- (1.8,0) node[midway,left] h; (2,0) -- (2,0.2) -- (1.8,0.2); [linecolor,line width=1.5pt] (0,0) -- (1.8,2) node[black,midway,above left] b -- (3,0) node[black,midway,above right] a -- cycle; [below] at (1.5,0) c; [below left] at (0,0) A; [below right] at (3,0) B; [above] at (1.8,2) C; tikzpicture — Figure 2.1.1Proof of the Law of Sines for an oblique triangle $\triangle\,ABC$

tikzpicture[every node/.style=font=] [fillcolor] (0,0) -- (5,2) -- (3,0) -- cycle; [dashed] (5.1,2.06) -- (5.1,0) node[midway,right] h -- (3,0); (4.9,0) -- (4.9,0.2) -- (5.1,0.2); [linecolor,line width=1.5pt] (0,0) -- (5,2) node[black,midway,above left] b -- (3,0) node[black,pos=0.7,above left] a -- cycle; [below] at (1.5,0) c; [below left] at (0,0) A; [below] at (3,0) B; [above] at (5,2) C; at (3.95,0.2) 180 - B; tikzpicture — Figure 2.1.1Proof of the Law of Sines for an oblique triangle $\triangle\,ABC$

Let $h$ be the height of the altitude. For each triangle in Figure 2.1.1, we see that

\begin{align} \frac{h}{b} ~&=~ \sin\;A\tag{2.4}\\ \intertext{and} \frac{h}{a} ~&=~ \sin\;B\tag{2.5}\\ \intertext{(in Figure \ref{fig:lawsines}(b), $\frac{h}{a} = \sin\;(180\Degrees - B) = \sin\;B$ by formula (\ref{eqn:sin180minus}) in Section 1.5). Thus, solving for $h$ in equation (\ref{eqn:hasinB}) and substituting that into equation (\ref{eqn:hbsinA}) gives} \frac{a\;\sin\;B}{b} ~&=~ \sin\;A ~,\tag{2.6}\\ \intertext{and so putting $a$ and $A$ on the left side and $b$ and $B$ on the right side, we get} \frac{a}{\sin\;A} ~&=~ \frac{b}{\sin\;B} ~.\tag{2.7}\\ \intertext{By a similar argument, drawing the altitude from $A$ to $\overline{BC}$ gives} \frac{b}{\sin\;B} ~&=~ \frac{c}{\sin\;C} ~, \tag{2.8}\end{align}

so putting the last two equations together proves the theorem. $\text{qed}$

Note that we did not prove the Law of Sines for right triangles, since it turns out (see Exercise 12) to be trivially true for that case.

Example 2.1

[r] tikzpicture[scale=0.75,every node/.style=font=] [fillcolor] (0,0) -- (1.8,2) -- (3,0) -- cycle; [linecolor,line width=1.5pt] (0,0) -- (1.8,2) node[black,midway,above left] b -- (3,0) node[black,midway,above right] a=10 -- cycle; [below] at (1.5,0) c; [below] at (0,0) A=41 ; [below right] at (3,0) B; [above] at (1.8,2) C=75 ; tikzpicture Case 1: One side and two angles.
Solve the triangle $\triangle\,ABC$ given $a = 10$, $A = 41\Degrees$, and $C = 75\Degrees$.

Solution: We can find the third angle by subtracting the other two angles from $180\Degrees$, then use the law of sines to find the two unknown sides. In this example we need to find $B$, $b$, and $c$. First, we see that

\begin{displaymath} B ~=~ 180\Degrees ~-~ A ~-~ C ~=~ 180\Degrees ~-~ 41\Degrees ~-~ 75\Degrees \quad\Rightarrow\quad \boxed{B ~=~ 64\Degrees} ~. \end{displaymath}

So by the Law of Sines we have

\begin{alignat*}{6} \frac{b}{\sin\;B} ~&=~ \frac{a}{\sin\;A} \quad&\Rightarrow\quad b ~&=~ \frac{a\;\sin\;B}{\sin\;A} ~&=~ \frac{10\;\sin\;64\Degrees}{\sin\;41\Degrees} \quad&\Rightarrow\quad \boxed{b ~=~ 13.7} ~,~\text{and}\\[4pt] \frac{c}{\sin\;C} ~&=~ \frac{a}{\sin\;A} \quad&\Rightarrow\quad c ~&=~ \frac{a\;\sin\;C}{\sin\;A} ~&=~ \frac{10\;\sin\;75\Degrees}{\sin\;41\Degrees} \quad&\Rightarrow\quad \boxed{c ~=~ 14.7} ~. \end{alignat*}

Example 2.2

[r] tikzpicture[scale=0.8,every node/.style=font=] [fillcolor] (0,0) -- (1.8,1.2) -- (3,0) -- cycle; [linecolor,line width=1.5pt] (0,0) -- (1.8,1.2) node[black,midway,above left] b=30 -- (3,0) node[black,midway,above right] a=18 -- cycle; [below] at (1.5,0) c; [below] at (0,0) A=25 ; [below right] at (3,0) B; [above] at (1.8,1.2) C; tikzpicture Case 2: Two sides and one opposite angle.
Solve the triangle $\triangle\,ABC$ given $a = 18$, $A = 25\Degrees$, and $b = 30$.

Solution: In this example we know the side $a$ and its opposite angle $A$, and we know the side $b$. We can use the Law of Sines to find the other opposite angle $B$, then find the third angle $C$ by subtracting $A$ and $B$ from $180\Degrees$, then use the law of sines to find the third side $c$. By the Law of Sines, we have

\begin{displaymath} \frac{\sin\;B}{b} ~=~ \frac{\sin\;A}{a} \quad\Rightarrow\quad \sin\;B ~=~ \frac{b\;\sin\;A}{a} ~=~ \frac{30\;\sin\;25\Degrees}{18} \quad\Rightarrow\quad \sin\;B ~=~ 0.7044 ~. \end{displaymath}

Using the 1pt $\sin^{-1}$ button on a calculator gives $B = 44.8\Degrees$. However, recall from Section 1.5 that $\sin\;(180\Degrees - B) = \sin\;B$. So there is a second possible solution for $B$, namely $180\Degrees - 44.8\Degrees = 135.2\Degrees$. Thus, we have to solve twice for $C$ and $c$ : once for $B = 44.8\Degrees$ and once for $B = 135.2\Degrees$:

$\boxed{B = 44.8\Degrees}$	$\boxed{B = 135.2\Degrees}$
$C = 180\Degrees - A - B = 180\Degrees - 25\Degrees - 44.8\Degrees = 110.2\Degrees$	$C = 180\Degrees - A - B = 180\Degrees - 25\Degrees - 135.2\Degrees = 19.8\Degrees$
$\dfrac{c}{\sin\;C} = \dfrac{a}{\sin\;A} ~\Rightarrow~ c = \dfrac{a\;\sin\;C}{\sin\;A} = \dfrac{18\;\sin\;110.2\Degrees}{\sin\;25\Degrees}$	$\dfrac{c}{\sin\;C} = \dfrac{a}{\sin\;A} ~\Rightarrow~ c = \dfrac{a\;\sin\;C}{\sin\;A} = \dfrac{18\;\sin\;19.8\Degrees}{\sin\;25\Degrees}$
$\Rightarrow~ c = 40$	$\Rightarrow~ c = 14.4$

Hence, $B = 44.8\Degrees$, $C = 110.2\Degrees$, $c = 40$ and $B = 135.2\Degrees$, $C = 19.8\Degrees$, $c = 14.4$ are the two possible sets of solutions. This means that there are two possible triangles, as shown in Figure 2.1.2.

tikzpicture[scale=0.8,every node/.style=font=] [fillcolor] (0,0) -- (1.8,1.2) -- (3,0) -- cycle; [linecolor,line width=1.5pt] (0,0) -- (1.8,1.2) node[black,midway,above left] b=30 -- (3,0) node[black,midway,above right] a=18 -- cycle; [below] at (1.5,0) c=40; [left] at (0,0) A=25 ; [right] at (3,0) B=44.8 ; [above] at (1.8,1.2) C=110.2 ; tikzpicture — Figure 2.1.2Two possible solutions

tikzpicture[scale=0.8,every node/.style=font=] [fillcolor] (0,0) -- (1.8,1.2) -- (1.2,0) -- cycle; [linecolor,line width=1.5pt] (0,0) -- (1.8,1.2) node[black,midway,above left] b=30 -- (1.2,0) node[black,pos=0.2,below right] a=18 -- cycle; [below] at (0.6,0) c=14.4; [left] at (0,0) A=25 ; [right] at (1.2,0) B=135.2 ; [above] at (1.8,1.2) C=19.8 ; tikzpicture — Figure 2.1.2Two possible solutions

In Example 2.2 we saw what is known as the ambiguous case. That is, there may be more than one solution. It is also possible for there to be exactly one solution or no solution at all.

Example 2.3

Case 2: Two sides and one opposite angle.
Solve the triangle $\triangle\,ABC$ given $a = 5$, $A = 30\Degrees$, and $b = 12$.

Solution: By the Law of Sines, we have

\begin{displaymath} \frac{\sin\;B}{b} ~=~ \frac{\sin\;A}{a} \quad\Rightarrow\quad \sin\;B ~=~ \frac{b\;\sin\;A}{a} ~=~ \frac{12\;\sin\;30\Degrees}{5} \quad\Rightarrow\quad \sin\;B ~=~ 1.2 ~, \end{displaymath}

which is impossible since $\abs{\sin\;B} \le 1$ for any angle $B$. Thus, there is no solution .

There is a way to determine how many solutions a triangle has in Case 2. For a triangle $\triangle\,ABC$, suppose that we know the sides $a$ and $b$ and the angle $A$. Draw the angle $A$ and the side $b$, and imagine that the side $a$ is attached at the vertex at $C$ so that it can “swing” freely, as indicated by the dashed arc in Figure 2.1.3 below.

tikzpicture[scale=1,every node/.style=font=] [dashed] (0,0) -- (3,0); [dashed] (1.8,1.2) -- (1.8,0) node[pos=0.4,left] h; (1.6,0) -- (1.6,0.2) -- (1.8,0.2); [dashed] ([shift=(1.8,1.2)] -35:0.8) arc (-35:-135:0.8); [linecolor,line width=1.5pt] (0,0) -- (1.8,1.2) node[black,midway,above left] b -- ++(-45:0.8) node[black,midway,above right] a; [below] at (0,0) A; [above] at (1.8,1.2) C; [below] at (2.5,0) B; tikzpicture — Figure 2.1.3The ambiguous case when $A$ is acute

tikzpicture[scale=1,every node/.style=font=] [white] (-0.5,0) -- (3.5,0); [dashed] (0,0) -- (3,0); [dashed] (1.8,1.2) -- (1.8,0) node[midway,left] h; (1.6,0) -- (1.6,0.2) -- (1.8,0.2); [dashed] ([shift=(1.8,1.2)] -50:1.2) arc (-50:-130:1.2); [linecolor,line width=1.5pt] (0,0) -- (1.8,1.2) node[black,midway,above left] b -- ++(-90:1.2) node[black,midway,right] a; [below] at (0.9,0) c; [below] at (0,0) A; [above] at (1.8,1.2) C; [below] at (1.8,0) B; tikzpicture — Figure 2.1.3The ambiguous case when $A$ is acute

If $A$ is acute, then the altitude from $C$ to $\overline{AB}$ has height $h = b\;\sin\;A$. As we can see in Figure 2.1.3(a)-(c), there is no solution when $a < h$ (this was the case in Example 2.3); there is exactly one solution - namely, a right triangle - when $a = h$; and there are two solutions when $h < a < b$ (as was the case in Example 2.2). When $a \ge b$ there is only one solution, even though it appears from Figure 2.1.3(d) that there may be two solutions, since the dashed arc intersects the horizontal line at two points. However, the point of intersection to the left of $A$ in Figure 2.1.3(d) can not be used to determine $B$, since that would make $A$ an obtuse angle, and we assumed that $A$ was acute.

If $A$ is not acute (i.e. $A$ is obtuse or a right angle), then the situation is simpler: there is no solution if $a \le b$, and there is exactly one solution if $a > b$ (see Figure 2.1.4).

tikzpicture[scale=1,every node/.style=font=] [dashed] (0,0) -- (2.5,0); [dashed] ([shift=(-0.5,1.2)] 0:1) arc (0:-85:1); [linecolor,line width=1.5pt] (0,0) -- (-0.5,1.2) node[black,midway,left] b -- ++(-20:1) node[black,midway,above right] a; [below] at (0,0) A; [above] at (-0.5,1.2) C; [below] at (2.2,0) B; tikzpicture — Figure 2.1.4The ambiguous case when $A \ge 90\Degrees$

tikzpicture[scale=1,every node/.style=font=] [dashed] (0,0) -- (2.7,0); [dashed] ([shift=(-0.5,1.2)] -5:2.955) arc (-5:-35:2.955); [linecolor,line width=1.5pt] (0,0) -- (-0.5,1.2) node[black,midway,left] b -- (2.2,0) node[black,midway,above right] a; [below] at (0,0) A; [above] at (-0.5,1.2) C; [below] at (2.3,0) B; tikzpicture — Figure 2.1.4The ambiguous case when $A \ge 90\Degrees$

Table 2.1 summarizes the ambiguous case of solving $\triangle\,ABC$ when given $a$, $A$, and $b$. Of course, the letters can be interchanged, e.g. replace $a$ and $A$ by $c$ and $C$, etc.

Table 2.1Summary of the ambiguous case

$0\Degrees < A < 90\Degrees$		$90\Degrees \le A < 180\Degrees$
$a < b\;\sin\;A$ :	No solution	$a \le b$ :	No solution
$a = b\;\sin\;A$ :	One solution	$a > b$ :	One solution
$b\;\sin\;A < a < b$ :	Two solutions
$a \ge b$ :	One solution

There is an interesting geometric consequence of the Law of Sines. Recall from Section 1.1 that in a right triangle the hypotenuse is the largest side. Since a right angle is the largest angle in a right triangle, this means that the largest side is opposite the largest angle. What the Law of Sines does is generalize this to any triangle:

In any triangle, the largest side is opposite the largest angle.

To prove this, let $C$ be the largest angle in a triangle $\triangle\,ABC$. If $C = 90\Degrees$ then we already know that its opposite side $c$ is the largest side. So we just need to prove the result for when $C$ is acute and for when $C$ is obtuse. In both cases, we have $A \le C$ and $B \le C$. We will first show that $\sin\;A \le \sin\;C$ and $\sin\;B \le \sin\;C$.

tikzpicture[every node/.style=font=] [latex-latex,black!60,line width=0.3pt] (0,2.5) node[above] y |- (3,0) node[right] x; [-latex] (0:1) arc (0:20:1); [linecolor,line width=1.5pt] (0,0) -- (20:2.5); [-latex] (0:1.7) arc (0:60:1.7); [dashed,linecolor,line width=1.5pt] (0,0) -- (60:2.5); at (15:2.1) r; at (65:1.9) r; (0,0) circle (2pt); (20:2.5) circle (2pt) node[right] (x_1,y_1);; (60:2.5) circle (2pt) node[above right] (x_2,y_2);; [dashed,-latex] (20:2.5) arc (20:59:2.5); at (11:1.23) A; at (42:1.9) C; tikzpicture — Figure 2.1.5

If $C$ is acute, then $A$ and $B$ are also acute. Since $A \le C$, imagine that $A$ is in standard position in the $xy$-coordinate plane and that we rotate the terminal side of $A$ counterclockwise to the terminal side of the larger angle $C$, as in Figure 2.1.5. If we pick points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ on the terminal sides of $A$ and $C$, respectively, so that their distance to the origin is the same number $r$, then we see from the picture that $y_{1} \le y_{2}$, and hence

\begin{displaymath} \sin\;A ~=~ \frac{y_{1}}{r} ~\le~ \frac{y_{2}}{r} ~=~ \sin\;C ~. \end{displaymath}

By a similar argument, $B \le C$ implies that $\sin\;B \le \sin\;C$. Thus, $\sin\;A \le \sin\;C$ and $\sin\;B \le \sin\;C$ when $C$ is acute. We will now show that these inequalities hold when $C$ is obtuse.

If $C$ is obtuse, then $180\Degrees - C$ is acute, as are $A$ and $B$. If $A > 180\Degrees - C$ then $A + C > 180\Degrees$, which is impossible. Thus, we must have $A \le 180\Degrees - C$. Likewise, $B \le 180\Degrees - C$. So by what we showed above for acute angles, we know that $\sin\;A \le \sin\;(180\Degrees - C)$ and $\sin\;B \le \sin\;(180\Degrees - C)$. But we know from Section 1.5 that $\sin\;C = \sin\;(180\Degrees - C)$. Hence, $\sin\;A \le \sin\;C$ and $\sin\;B \le \sin\;C$ when $C$ is obtuse.

Thus, $\sin\;A \le \sin\;C$ if $C$ is acute or obtuse, so by the Law of Sines we have

\begin{align*} \frac{a}{c} ~=~ \frac{\sin\;A}{\sin\;C} ~\le~ \frac{\sin\;C}{\sin\;C} ~=~ 1 \quad\Rightarrow\quad \frac{a}{c} ~\le~ 1 \quad\Rightarrow\quad a ~\le~ c ~. \end{align*}

By a similar argument, $b \le c$. Thus, $a \le c$ and $b \le c$, i.e. $c$ is the largest side. $\text{qed}$

Exercises

For Exercises 1-9, solve the triangle $\triangle\,ABC$.

$a = 10$, $A = 35\Degrees$, $B = 25\Degrees$
$b = 40$, $B = 75\Degrees$, $c = 35$
$A = 40\Degrees$, $B = 45\Degrees$, $c = 15$

3
$a = 5$, $A = 42\Degrees$, $b = 7$
$a = 40$, $A = 25\Degrees$, $c = 30$
$a = 5$, $A = 47\Degrees$, $b = 9$

3
$a = 12$, $A = 94\Degrees$, $b = 15$
$a = 15$, $A = 94\Degrees$, $b = 12$
$a = 22$, $A = 50\Degrees$, $c = 27$
Draw a circle with a radius of $2$ inches and inscribe a triangle inside the circle. Use a ruler and a protractor to measure the sides $a$, $b$, $c$ and the angles $A$, $B$, $C$ of the triangle. The Law of Sines says that the ratios $\frac{a}{\sin\;A}$, $\frac{b}{\sin\;B}$, $\frac{c}{\sin\;C}$ are equal. Verify this for your triangle. What relation does that common ratio have to the diameter of your circle?
An observer on the ground measures an angle of inclination of $30\Degrees$ to an approaching airplane, and $10$ seconds later measures an angle of inclination of $55\Degrees$. If the airplane is flying at a constant speed and at a steady altitude of $6000$ ft in a straight line directly over the observer, find the speed of the airplane in miles per hour. (Note: $1$ mile = $5280$ ft)
Prove the Law of Sines for right triangles. (Hint: One of the angles is known.)
For a triangle $\triangle\,ABC$, show that $~\dfrac{a \pm b}{c} ~=~ \dfrac{\sin\;A \;\pm\; \sin\;B}{\sin\;C}\,$.
For a triangle $\triangle\,ABC$, show that $~\dfrac{a}{c} ~=~\dfrac{\sin\;(B+C)}{\sin\;C}\,$.
One diagonal of a parallelogram is 17 cm long and makes angles of $36\Degrees$ and $15\Degrees$ with the sides. Find the lengths of the sides.
Explain why in Case 1 (one side and two angles) there is always exactly one solution.

Recall from geometry that an altitude of a triangle is a perpendicular line segment from any vertex to the line containing the side opposite the vertex. ↩

\(\boxed{B = 44.8\Degrees}\)	\(\boxed{B = 135.2\Degrees}\)
\(C = 180\Degrees - A - B = 180\Degrees - 25\Degrees - 44.8\Degrees = 110.2\Degrees\)	\(C = 180\Degrees - A - B = 180\Degrees - 25\Degrees - 135.2\Degrees = 19.8\Degrees\)
\(\dfrac{c}{\sin\;C} = \dfrac{a}{\sin\;A} ~\Rightarrow~ c = \dfrac{a\;\sin\;C}{\sin\;A} = \dfrac{18\;\sin\;110.2\Degrees}{\sin\;25\Degrees}\)	\(\dfrac{c}{\sin\;C} = \dfrac{a}{\sin\;A} ~\Rightarrow~ c = \dfrac{a\;\sin\;C}{\sin\;A} = \dfrac{18\;\sin\;19.8\Degrees}{\sin\;25\Degrees}\)
\(\Rightarrow~ c = 40\)	\(\Rightarrow~ c = 14.4\)

\(0\Degrees < A < 90\Degrees\)		\(90\Degrees \le A < 180\Degrees\)
\(a < b\;\sin\;A\) :	No solution	\(a \le b\) :	No solution
\(a = b\;\sin\;A\) :	One solution	\(a > b\) :	One solution
\(b\;\sin\;A < a < b\) :	Two solutions
\(a \ge b\) :	One solution