[latex]\begin{gathered} \frac{\sin \alpha }{a}=\frac{\sin \beta }{b} \\ \frac{\sin \left(35^\circ \right)}{6}=\frac{\sin \beta }{8}\\ \frac{8\sin \left(35^\circ \right)}{6}=\sin \beta \\ 0.7648\approx \sin \beta \\ {\sin }^{-1}\left(0.7648\right)\approx 49.9^\circ \\ \beta \approx 49.9^\circ \end{gathered}[/latex]

However, in the diagram, angle [latex]\beta[/latex] appears to be an obtuse angle and may be greater than 90°. How did we get an acute angle, and how do we find the measurement of [latex]\beta ?[/latex] Let’s investigate further. Dropping a perpendicular from [latex]\gamma[/latex] and viewing the triangle from a right angle perspective, it appears that there may be a second triangle that will fit the given criteria.

The angle supplementary to [latex]\beta[/latex] is approximately equal to 49.9°, which means that [latex]\beta =180^\circ -49.9^\circ =130.1^\circ[/latex]. (Remember that the sine function is positive in both the first and second quadrants.) Solving for [latex]\gamma[/latex], we have

[latex]\gamma =180^\circ -35^\circ -130.1^\circ \approx 14.9^\circ[/latex]

We can then use these measurements to solve the other triangle. Since [latex]{\gamma }^{\prime }[/latex] is supplementary to [latex]\gamma[/latex], we have

[latex]{\gamma }^{\prime }=180^\circ -35^\circ -49.9^\circ \approx 95.1^\circ[/latex]

Now we need to find [latex]c[/latex] and [latex]{c}^{\prime }[/latex].

We have

[latex]\begin{gathered}\frac{c}{\sin \left(14.9^\circ \right)}=\frac{6}{\sin \left(35^\circ \right)} \\ c=\frac{6\sin \left(14.9^\circ \right)}{\sin \left(35^\circ \right)}\approx 2.7 \end{gathered}[/latex]

Finally,

[latex]\begin{gathered}\frac{{c}^{\prime }}{\sin \left(95.1^\circ \right)}=\frac{6}{\sin \left(35^\circ \right)} \\ {c}^{\prime }=\frac{6\sin \left(95.1^\circ \right)}{\sin \left(35^\circ \right)}\approx 10.4 \end{gathered}[/latex]

To summarize, there are two triangles with an angle of 35°, an adjacent side of 8, and an opposite side of 6.

However, we were looking for the values for the triangle with an obtuse angle [latex]\beta[/latex]. We can see them in the first triangle (a) in Figure 12.

[latex]\begin{align}\frac{\sin \left(85^\circ \right)}{12}&=\frac{\sin \beta }{9} && \text{Isolate the unknown}.\\ \frac{9\sin \left(85^\circ \right)}{12}&=\sin \beta \end{align}[/latex]

To find [latex]\beta[/latex], apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for [latex]\beta[/latex]. It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions.

[latex]\begin{align}\beta &={\sin }^{-1}\left(\frac{9\sin \left(85^\circ \right)}{12}\right) \\ \beta &\approx {\sin }^{-1}\left(0.7471\right) \\ \beta &\approx 48.3^\circ \end{align}[/latex]

In this case, if we subtract [latex]\beta[/latex] from 180°, we find that there may be a second possible solution. Thus, [latex]\beta =180^\circ -48.3^\circ \approx 131.7^\circ[/latex]. To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives

[latex]\alpha =180^\circ -85^\circ -131.7^\circ \approx -36.7^\circ[/latex],

which is impossible, and so [latex]\beta \approx 48.3^\circ[/latex].

To find the remaining missing values, we calculate [latex]\alpha =180^\circ -85^\circ -48.3^\circ \approx 46.7^\circ[/latex]. Now, only side [latex]a[/latex] is needed. Use the Law of Sines to solve for [latex]a[/latex] by one of the proportions.

[latex]\begin{gathered} \frac{\sin \left(85^\circ \right)}{12}=\frac{\sin \left(46.7^\circ \right)}{a} \\ a\frac{\sin \left(85^\circ \right)}{12}=\sin \left(46.7^\circ \right) \\ a=\frac{12\sin \left(46.7^\circ \right)}{\sin \left(85^\circ \right)}\approx 8.8 \end{gathered}[/latex]

The complete set of solutions for the given triangle is

[latex]\begin{gathered} \alpha \approx 46.7^\circ \text{, }a\approx 8.8 \\ \beta \approx 48.3^\circ \text{, }b=9 \\ \gamma =85^\circ \text{, }c=12\end{gathered}[/latex]

[latex]\begin{gathered}\frac{\sin \alpha }{10}=\frac{\sin \left(50^\circ \right)}{4} \\ \sin \alpha =\frac{10\sin \left(50^\circ \right)}{4} \\ \sin \alpha \approx 1.915 \end{gathered}[/latex]

We can stop here without finding the value of [latex]\alpha[/latex]. Because the range of the sine function is [latex]\left[-1,1\right][/latex], it is impossible for the sine value to be 1.915. In fact, inputting [latex]{\sin }^{-1}\left(1.915\right)[/latex] in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions.