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Section 4.6 The Law of Sines and the Law of Cosines

Live your life as though your every act were to become a universal law. — Immanuel Kant, German philosopher, 1724 – 1804

Problem: Evaluate \(x\text{.}\) See Figure 4.61.

Figure 4.61. The unknown length \(x\text{.}\)

Notation: Vertices, Sides, and Angles of a Triangle See Figure 4.62.

Figure 4.62. Vertices, Sides, and Angles

In a triangle \(\triangle ABC\) with vertices \(A\text{,}\) \(B\text{,}\) and \(C\text{:}\)

  • \(\alpha\) is the angle at \(A\text{.}\)

  • \(a\) is the side opposite \(A\) and \(\alpha\text{.}\)

  • \(\beta\) is the angle at \(B\text{.}\)

  • \(b\) is the side opposite \(B\) and \(\beta\text{.}\)

  • \(\gamma\) is the angle at \(C\text{.}\)

  • \(c\) is the side opposite \(C\) and \(\gamma\text{.}\)

The Law of Sines.

Figure 4.63. Law of sines

We use Figure 4.63.

Observe that

\begin{equation*} \sin \alpha = \frac{h}{b} \text{ and } \sin \beta = \frac{h}{a}. \end{equation*}

Now we calculate the length of \(h\) in two different ways:

\begin{equation*} h = b\sin\alpha \text{ and } h = a\sin\beta. \end{equation*}

Hence

\begin{equation*} b \sin \alpha = a \sin \beta\text{.} \end{equation*}

Similarly, \(c \sin \alpha = a \sin \gamma\text{.}\)

We write

\begin{equation*} D = \frac{a}{\sin \alpha}\text{.} \end{equation*}

Since \(b \sin \alpha = a \sin \beta\) and \(c \sin \alpha = a \sin \gamma\text{,}\) it follows that

\begin{equation*} D = \frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma}. \end{equation*}

This is known as the Law of Sines.

Example 4.6.1. Law of sines.

In the triangle \(\triangle ABC\) with the usual notation, suppose that \(\sin \alpha = \frac{2\sqrt{2}}{3}\text{,}\) that \(a = 11\) and that \(b = 9\text{.}\) Find \(\sin \beta\text{.}\)

Solution

By the Law of Sines

\begin{equation*} \frac{a}{\sin \alpha} = \frac{b}{\sin \beta} \Rightarrow \frac{11}{\frac{2\sqrt{2}}{3}} = \frac{9}{\sin \beta}\Rightarrow \sin\beta =\frac{9\cdot\frac{2\sqrt{2}}{3}}{11} = \frac{6\sqrt{2}}{11}. \end{equation*}

Area of a Triangle: The Law of Sines implies that

\begin{equation*} \text{Area}(\triangle ABC) = \frac{1}{2} ac \sin \beta. \end{equation*}

Solution

See Figure 4.64.

Figure 4.64. Area of \(\triangle ABC\text{.}\)

In \(\triangle ABC\) with the usual notation, let \(h\) be the hight to the side \(c\text{.}\)

Observe that \(h=a\sin \beta\text{.}\) Thus, the area of the triangle is given by

\begin{equation*} A=\frac{ch}{2}=\frac{1}{2}\cdot ac\sin\beta . \end{equation*}

Note: Notice that in this formula, \(\beta\) represents the anglebetween \(a\) and \(c\text{.}\)

Example 4.6.2. Use the given area to find a side.

In \(\triangle ABC\) with the usual notation, suppose that \(\sin \alpha = 2/3\text{,}\) that \(b = 9\) and that the area of \(\triangle ABC\) is \(24\) Find \(c\text{.}\)

Solution

Note that, in the usual notation, \(\alpha\) represents the angle between \(b\) and \(c\text{,}\) so the area of \(\triangle ABC\) is given by

\begin{equation*} A=\frac{1}{2}\cdot bc\sin\alpha\Leftrightarrow c=\frac{2A}{b\sin \alpha}. \end{equation*}

We conclude that in the given triangle

\begin{equation*} c=\frac{2\cdot 24}{9\cdot \frac{2}{3}}=\frac{2\cdot 24}{6}=8. \end{equation*}

The Law of Cosines. See Figure 4.65.

See Figure 4.65.

Figure 4.65. Area of \(\triangle ABC\text{.}\)

We use Figure 4.65.

Observe that, by the Pythagorean theorem,

\begin{equation*} b^2 =(c-a\cos\beta)^2+(a\sin\beta)^2= c^2-2ac\cos\beta+a^2\cos^2\beta+a^2\sin^2\beta \end{equation*}
\begin{equation*} =c^2-2ac\cos\beta+a^2(\cos^2\beta+\sin^2\beta)=c^2-2ac\cos\beta+a^2. \end{equation*}

The fact that

\begin{equation*} b^2 =a^2+c^2-2ac\cos\beta \end{equation*}

is known as the Law of Cosines.

Note: Observe that the the angle \(\beta\) is between the sides \(a\) and \(c\) and that side \(b\) is opposite to the angle \(\beta\text{.}\) Hence

\begin{equation*} c^2 =a^2+b^2-2ab\cos\gamma \text{ and } a^2 =b^2+c^2-2bc\cos\alpha \end{equation*}
Example 4.6.3. Solve using the Law of Cosines.

Evaluate \(x\text{.}\) See Figure 4.66.

Figure 4.66. The unknown length \(x\text{.}\)
Solution

By the Law of Cosines,

\begin{equation*} 5.8^2 =x^2+4^2-2\cdot 4\cdot x\cdot \cos \frac{4\pi}{9}\Leftrightarrow x^2-8x\cos\frac{4\pi}{9}+4^2-5.8^2=x^2-8x\cos\frac{4\pi}{9}-17.64=0. \end{equation*}

We solve the quadratic equation to obtain

\begin{equation*} x=\frac{8\cos\frac{4\pi}{9}\pm\sqrt{64\cos\frac{4\pi}{9}+70.26}}{2}\approx\frac{1.39 \pm9.02}{2} \end{equation*}

and, since \(x\gt 0\) (why?), we reject the negative option.

Thus

\begin{equation*} x\approx\frac{1.39+9.02}{2}=5.215. \end{equation*}