10.1 Non-right triangles: law of sines  (Page 3/10)

In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle $\gamma =\mathrm{85°},$ and its corresponding side $c=12,$ and we know side $b=9.$ We will use this proportion to solve for $\beta .$

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

In this case, if we subtract $\beta$ from 180°, we find that there may be a second possible solution. Thus, $\beta =\mathrm{180°}-\mathrm{48.3°}\approx \mathrm{131.7°}.$ To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives

which is impossible, and so $\beta \approx \mathrm{48.3°}.$

To find the remaining missing values, we calculate $\alpha =\mathrm{180°}-\mathrm{85°}-\mathrm{48.3°}\approx \mathrm{46.7°}.$ Now, only side $a$ is needed. Use the Law of Sines to solve for $a$ by one of the proportions.

The complete set of solutions for the given triangle is

Using the given information, we can solve for the angle opposite the side of length 10. See [link] .

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