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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° .


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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, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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