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Solving for two unknown sides and angle of an aas triangle

Solve the triangle shown in [link] to the nearest tenth.

An oblique triangle with standard labels. Angle alpha is 50 degrees, angle gamma is 30 degrees, and side a is of length 10. Side b is the horizontal base.

The three angles must add up to 180 degrees. From this, we can determine that

β = 180° 50° 30° = 100°

To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle α = 50° and its corresponding side a = 10. We can use the following proportion from the Law of Sines to find the length of c .

sin ( 50° ) 10 = sin ( 30° ) c c sin ( 50° ) 10 = sin ( 30° ) Multiply both sides by  c . c = sin ( 30 ° ) 10 sin ( 50° ) Multiply by the reciprocal to isolate  c . c 6.5

Similarly, to solve for b , we set up another proportion.

    sin ( 50° ) 10 = sin ( 100° ) b    b sin ( 50° ) = 10 sin ( 100° ) Multiply both sides by  b .                 b = 10 sin ( 100° ) sin ( 50° ) Multiply by the reciprocal to isolate  b .                 b 12.9

Therefore, the complete set of angles and sides is

α = 50° a = 10 β = 100° b 12.9 γ = 30° c 6.5

Solve the triangle shown in [link] to the nearest tenth.

An oblique triangle with standard labels. Angle alpha is 98 degrees, angle gamma is 43 degrees, and side b is of length 22. Side b is the horizontal base.

α = 98 a = 34.6 β = 39 b = 22 γ = 43 c = 23.8

Using the law of sines to solve ssa triangles

We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an ambiguous case    . Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution.

Possible outcomes for ssa triangles

Oblique triangles in the category SSA may have four different outcomes. [link] illustrates the solutions with the known sides a and b and known angle α .

Four attempted oblique triangles are in a row, all with standard labels. Side c is the horizontal base. In the first attempted triangle, side a is less than the altitude height. Since side a cannot reach side c,  there is no triangle. In the second attempted triangle, side a is equal to the length of the altitude height, so side a forms a right angle with side c. In the third attempted triangle, side a is greater than the altitude height and less than side b, so side a can form either an acute or obtuse angle with side c. In the fourth attempted triangle, side a is greater than or equal to side b, so side a forms an acute angle with side c.

Solving an oblique ssa triangle

Solve the triangle in [link] for the missing side and find the missing angle measures to the nearest tenth.

An oblique triangle with standard labels where side a is of length 6, side b is of length 8, and angle alpha is 35 degrees.

Use the Law of Sines to find angle β and angle γ , and then side c . Solving for β , we have the proportion

sin α a = sin β b sin ( 35° ) 6 = sin β 8 8 sin ( 35° ) 6 = sin β 0.7648 sin β sin 1 ( 0.7648 ) 49.9° β 49.9°

However, in the diagram, angle β appears to be an obtuse angle and may be greater than 90°. How did we get an acute angle, and how do we find the measurement of β ? Let’s investigate further. Dropping a perpendicular from γ and viewing the triangle from a right angle perspective, we have [link] . It appears that there may be a second triangle that will fit the given criteria.

An oblique triangle built from the previous with standard prime labels. Side a is of length 6, side b is of length 8, and angle alpha prime is 35 degrees. An isosceles triangle is attached, using side a as one of its congruent legs and the angle supplementary to angle beta as one of its congruent base angles. The other congruent angle is called beta prime, and the entire new horizontal base, which extends from the original side c, is called c prime. There is a dotted altitude line from angle gamma prime to side c prime.

The angle supplementary to β is approximately equal to 49.9°, which means that β = 180° 49.9° = 130.1° . (Remember that the sine function is positive in both the first and second quadrants.) Solving for γ , we have

γ = 180° 35° 130.1° 14.9°

We can then use these measurements to solve the other triangle. Since γ is supplementary to γ , we have

γ = 180° 35° 49.9° 95.1°

Now we need to find c and c .

We have

c sin ( 14.9° ) = 6 sin ( 35° )                c = 6 sin ( 14.9° ) sin ( 35° ) 2.7


c sin ( 95.1° ) = 6 sin ( 35° )               c = 6 sin ( 95.1° ) sin ( 35° ) 10.4

To summarize, there are two triangles with an angle of 35°, an adjacent side of 8, and an opposite side of 6, as shown in [link] .

There are two triangles with standard labels. Triangle a is the orginal triangle. It has angles alpha of 35 degrees, beta of 130.1 degrees, and gamma of 14.9 degrees. It has sides a = 6, b = 8, and c is approximately 2.7. Triangle b is the extended triangle. It has angles alpha prime = 35 degrees, angle beta prime = 49.9 degrees, and angle gamma prime = 95.1 degrees. It has side a prime = 6, side b prime = 8, and side c prime is approximately 10.4.

However, we were looking for the values for the triangle with an obtuse angle β . We can see them in the first triangle (a) in [link] .

Questions & Answers

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Source:  OpenStax, Contemporary math applications. OpenStax CNX. Dec 15, 2014 Download for free at http://legacy.cnx.org/content/col11559/1.6
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