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

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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

α = 80° a = 120 β ≈ 83.2° b = 121 γ ≈ 16.8° c ≈ 35.2

Solution 2

α ′ = 80°                                   a ′ = 120 β ′ ≈ 96.8°                           b ′ = 121 γ ′ ≈ 3.2°                                 c ′ ≈ 6.8
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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 .

      sin ( 85° ) 12 = sin   β 9 Isolate the unknown .   9 sin ( 85° ) 12 = sin   β

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.

β = sin − 1 ( 9 sin ( 85° ) 12 ) β ≈ sin − 1 ( 0.7471 ) β ≈ 48.3°

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

α = 180° − 85° − 131.7° ≈ − 36.7° ,

which is impossible, and so   β ≈ 48.3° .

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

  sin ( 85 ° ) 12 = sin ( 46.7 ° ) a   a sin ( 85 ° ) 12 = sin ( 46.7 ° )                         a = 12 sin ( 46.7 ° ) sin ( 85 ° ) ≈ 8.8

The complete set of solutions for the given triangle is

α ≈ 46.7°         a ≈ 8.8 β ≈ 48.3°         b = 9 γ = 85°             c = 12
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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 ≈ 5.7° , γ ≈ 94.3° , c ≈ 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] .

  sin   α 10 = sin ( 50° ) 4     sin   α = 10 sin ( 50° ) 4     sin   α ≈ 1.915
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 = 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 = opposite hypotenuse   to write an equation for area in oblique triangles. In the acute triangle, we have   sin   α = 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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