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Solution.

The altitude of the triangle is

\(h=b\sin A=7\sin40^{\circ}\approx4.50\text{.}\) Since

\(A\) is an acute angle and the altitude (

\(h\approx4.50\)) is less than the opposite side (

\(a=6\)) which is less than the adjacent side (

\(b=7)\text{.}\) This satisfies Case 3 of

Table 4.1.7, thus there are two triangles.

Using the first form for the Law of Sines

\begin{equation*}
\frac{\sin A}{a}=\frac{\sin B}{b}
\end{equation*}

we get

\begin{equation*}
\sin B=\frac{b\sin A}{a}=\frac{7\sin40^{\circ}}{6}\approx0.75
\end{equation*}

Here we have that \(\sin B\) is positive and we know that sine is positive in Quadrant I and II so our value for \(B\) may be between \(0^{\circ}\) and \(180^{\circ}\text{.}\) Taking the inverse sine we get our first angle

\begin{equation*}
B_1=\sin^{-1}(0.75)\approx48.58^{\circ}
\end{equation*}

which is in Quadrant I. To calculate the angle in Quadrant II, we subtract the reference angle from \(180^{\circ}\) to get

\begin{equation*}
B_2=180^{\circ}-B_1=180^{\circ}-48.58^{\circ}=131.42^{\circ}
\end{equation*}

So we now have two triangles: \(AB_1C\) and \(AB_2C\text{.}\)

To solve triangle \(AB_1C\text{,}\) be begin by finding \(C_1\)

\begin{equation*}
C_1\approx180^{\circ}-A-B_1=180^{\circ}-40^{\circ}-48.58^{\circ}=91.42^{\circ}
\end{equation*}

To find the side \(c_1\text{,}\) we use the second form for the Law of Sines

\begin{equation*}
\frac{c_1}{\sin C}=\frac{a}{\sin A}
\end{equation*}

Thus

\begin{equation*}
c_1=\frac{a\sin C_1}{\sin A}\approx\frac{6\sin48.58^{\circ}}{\sin40^{\circ}}\approx9.33
\end{equation*}

To solve triangle \(AB_2C\text{,}\) we begin by finding \(C_2\)

\begin{equation*}
C_2\approx180^{\circ}-A-B_2=180^{\circ}-40^{\circ}-131.42^{\circ}=8.58^{\circ}
\end{equation*}

To find the side \(c_2\text{,}\) we use the second form for the Law of Sines to get

\begin{equation*}
c_2=\frac{a\sin C_2}{\sin A}\approx\frac{6\sin8.58^{\circ}}{\sin40^{\circ}}\approx1.39
\end{equation*}