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

Precalculus

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Given $α = 80°, a = 120,$ and $b = 121,$ find the missing side and angles. If there is more than one possible solution, show both.

Solution 1

\begin{array}{l} α = 80° & a = 120 \\ β \approx 83.2° & b = 121 \\ γ \approx 16.8° & c \approx 35.2 \end{array}

Solution 2

\begin{array}{l} α^{'} = 80° a^{'} = 120 \\ β^{'} \approx 96.8° b^{'} = 121 \\ γ^{'} \approx 3.2° c^{'} \approx 6.8 \end{array}

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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.

An oblique triangle with standard labels. Side b is 9, side c is 12, and angle gamma is 85. Angle alpha, angle beta, and side a are unknown.

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 $γ = 85°,$ and its corresponding side $c = 12,$ and we know side $b = 9.$ We will use this proportion to solve for $β .$

\begin{array}{l} \frac{\sin (85°)}{12} = \frac{\sin β}{9} \begin{matrix}  \end{matrix} & Isolate the unknown . \\ \frac{9 \sin (85°)}{12} = \sin β \end{array}

To find $β,$ apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for $β .$ 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} β = \sin^{- 1} (\frac{9 \sin (85°)}{12}) \\ β \approx \sin^{- 1} (0.7471) \\ β \approx 48.3° \end{array}

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

α = 180° - 85° - 131.7° \approx - 36.7°,

which is impossible, and so $β \approx 48.3° .$

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

\begin{array}{l} \begin{array}{l} \begin{array}{l} \frac{\sin (85 °)}{12} = \frac{\sin (46.7 °)}{a} \end{array} \end{array} \\ a \frac{\sin (85 °)}{12} = \sin (46.7 °) \\ a = \frac{12 \sin (46.7 °)}{\sin (85 °)} \approx 8.8 \end{array}

The complete set of solutions for the given triangle is

\begin{array}{l} \begin{array}{l} α \approx 46.7° a \approx 8.8 \end{array} \\ β \approx 48.3° b = 9 \\ γ = 85° c = 12 \end{array}

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Given $α = 80°, a = 100, b = 10,$ find the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth.

$β \approx 5.7°, γ \approx 94.3°, c \approx 101.3$

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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} \frac{\sin α}{10} = \frac{\sin (50°)}{4} \\ \sin α = \frac{10 \sin (50°)}{4} \\ \sin α \approx 1.915 \end{array}

An incomplete triangle. One side has length 4 opposite a 50 degree angle, and a second side has length 10 opposite angle a. The side of length 4 is too short to reach the side of length 10, so there is no third angle.

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

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Determine the number of triangles possible given $a = 31, b = 26, β = 48° .$

two

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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 $Area = \frac{1}{2} b h,$ where $b$ is base and $h$ is height. For oblique triangles, we must find $h$ 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 $\sin α = \frac{opposite}{hypotenuse}$ to write an equation for area in oblique triangles. In the acute triangle, we have $\sin α = \frac{h}{c}$ or $c \sin α = h .$ However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base $b$ to form a right triangle. The angle used in calculation is $α^{'},$ or $180 - α .$

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Source: OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6

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