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

 Page 3 / 10

Given $\text{\hspace{0.17em}}\alpha =80°,a=120,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b=121,\text{\hspace{0.17em}}$ find the missing side and angles. If there is more than one possible solution, show both.

Solution 1

$\begin{array}{ll}\alpha =80°\hfill & a=120\hfill \\ \beta \approx 83.2°\hfill & b=121\hfill \\ \gamma \approx 16.8°\hfill & c\approx 35.2\hfill \end{array}$

Solution 2

$\begin{array}{l}{\alpha }^{\prime }=80°\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{a}^{\prime }=120\hfill \\ {\beta }^{\prime }\approx 96.8°\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{b}^{\prime }=121\hfill \\ {\gamma }^{\prime }\approx 3.2°\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{c}^{\prime }\approx 6.8\hfill \end{array}$

## Solving for the unknown sides and angles of a ssa triangle

In the triangle shown in [link] , solve for the unknown side and angles. Round your answers to the nearest tenth.

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 $\text{\hspace{0.17em}}\gamma =85°,\text{\hspace{0.17em}}$ and its corresponding side $\text{\hspace{0.17em}}c=12,\text{\hspace{0.17em}}$ and we know side $\text{\hspace{0.17em}}b=9.\text{\hspace{0.17em}}$ We will use this proportion to solve for $\text{\hspace{0.17em}}\beta .$

To find $\text{\hspace{0.17em}}\beta ,\text{\hspace{0.17em}}$ apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for $\text{\hspace{0.17em}}\beta .\text{\hspace{0.17em}}$ It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions.

$\begin{array}{l}\beta ={\mathrm{sin}}^{-1}\left(\frac{9\mathrm{sin}\left(85°\right)}{12}\right)\hfill \\ \beta \approx {\mathrm{sin}}^{-1}\left(0.7471\right)\hfill \\ \beta \approx 48.3°\hfill \end{array}$

In this case, if we subtract $\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$ from 180°, we find that there may be a second possible solution. Thus, $\text{\hspace{0.17em}}\beta =180°-48.3°\approx 131.7°.\text{\hspace{0.17em}}$ To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives

$\alpha =180°-85°-131.7°\approx -36.7°,$

which is impossible, and so $\text{\hspace{0.17em}}\beta \approx 48.3°.$

To find the remaining missing values, we calculate $\text{\hspace{0.17em}}\alpha =180°-85°-48.3°\approx 46.7°.\text{\hspace{0.17em}}$ Now, only side $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is needed. Use the Law of Sines to solve for $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ by one of the proportions.

The complete set of solutions for the given triangle is

Given $\text{\hspace{0.17em}}\alpha =80°,a=100,\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=10,\text{\hspace{0.17em}}$ find the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth.

$\beta \approx 5.7°,\gamma \approx 94.3°,c\approx 101.3$

## Finding the triangles that meet the given criteria

Find all possible triangles if one side has length 4 opposite an angle of 50°, and a second side has length 10.

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

$\begin{array}{l}\text{\hspace{0.17em}}\frac{\mathrm{sin}\text{\hspace{0.17em}}\alpha }{10}=\frac{\mathrm{sin}\left(50°\right)}{4}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =\frac{10\mathrm{sin}\left(50°\right)}{4}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha \approx 1.915\hfill \end{array}$

We can stop here without finding the value of $\text{\hspace{0.17em}}\alpha .\text{\hspace{0.17em}}$ Because the range of the sine function is $\text{\hspace{0.17em}}\left[-1,1\right],\text{\hspace{0.17em}}$ it is impossible for the sine value to be 1.915. In fact, inputting $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(1.915\right)\text{\hspace{0.17em}}$ in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions.

Determine the number of triangles possible given $\text{\hspace{0.17em}}a=31,\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=26,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\beta =48°.\text{\hspace{0.17em}}\text{\hspace{0.17em}}$

two

## Finding the area of an oblique triangle using the sine function

Now that we can solve a triangle for missing values, we can use some of those values and the sine function to find the area of an oblique triangle. Recall that the area formula for a triangle is given as $\text{\hspace{0.17em}}\text{Area}=\frac{1}{2}bh,\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ is base and $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ is height. For oblique triangles, we must find $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ before we can use the area formula. Observing the two triangles in [link] , one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =\frac{\text{opposite}}{\text{hypotenuse}}\text{\hspace{0.17em}}$ to write an equation for area in oblique triangles. In the acute triangle, we have $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =\frac{h}{c}\text{\hspace{0.17em}}$ or $c\mathrm{sin}\text{\hspace{0.17em}}\alpha =h.\text{\hspace{0.17em}}$ However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ to form a right triangle. The angle used in calculation is $\text{\hspace{0.17em}}{\alpha }^{\prime },\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}180-\alpha .$

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