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Ambiguous Triangles
Given triangular parts SSS , ASA or AAS always guarantees a single, unique triangle. Given the triangular parts SSA , however, is different and leaves the triangle unclear, or ambiguous. The “ Ambiguous Case ” (SSA) occurs when we are given two sides and the angle opposite one of these given sides. The triangles resulting from this condition needs to be explored much more closely than the SSS, ASA, and AAS cases, for SSA may result in one triangle, two triangles, or even no triangle at all!
Solving Triangles – The Ambiguous Case (SSA)
Given: angle A sides a, b
Zero Triangles
One Triangle
Two Triangles
The given angle A is greater than 90
angle A > 90
a b a > b
The given angle A is less than 90 angle A < 90
a < h or a < b
h (height of triangle) = b sin A
a = h or a b
h < a or a < b
Law of Sines 𝑎 sin 𝐴
sin 𝐵
sin 𝐶
When relying on a < b, must then use the Law of Sines to determine sin B
If sin B > 1, NO TRIANGLES
If sin B < 1, TWO TRIANGLES (found in quadrants I and II)
Two Examples
Solving Triangles for the Ambiguous Case (SSA)
Example #1 (No Triangles)
Given A = 42, a = 3, b = 8
Since A = 42 < 90 and a < b, we calculate the value of sin B using the Law of Sines:
3 sin 42° =^
8 sin 𝐵 yields that sin^ B^ = 1.784 which is greater than one
(recall that -1 < sin B < +1). Hence, there are no possible triangles and nothing to solve for.
Example #2 (Two Triangles)
Given A = 34, a = 2, b = 3
Since A = 34 < 90 and a < b, we again calculate the value of sin B using the Law of Sines:
2 sin 34° =^
3 sin 𝐵 yields that sin^ B^ = 0.839, which is between zero and one. Hence there will be two possible triangles to solve for.
First Triangle
𝑩𝟏 = sin−1^ 0.839 = 57.01
Therefore 𝑪𝟏 = 180° − 34° − 57.01° = 88.99°
Finally, again using the Law of Sines, 2 sin 34°
𝑐 sin 88.99° and, solving this equation for c, we get 𝑐 1 = 3.58.
Second Triangle
Angle 𝑩𝟐 is found by subtracting < 𝑩𝟏 from 180.
Thus < 𝑩𝟐 = 180 − 57.01 = 122.99°.
Angle 𝑪𝟐 = 180° − 34° − 122.99° = 23.01°
And to find the missing side, 𝑐 2 , we solve the ratio
2 sin 34°
𝑐 2 sin 23.01°
Hence, side 𝑐 2 = 1.40.
Triangle # < 𝐵 1 = 57 .01° < 𝐶 1 = 88 .99° 𝑠𝑖𝑑𝑒 𝑐 1 = 3. 58
Triangle # < 𝐵 2 = 122 .99° < 𝐶 2 = 23 .01° 𝑠𝑖𝑑𝑒 𝑐 2 = 1. 40
FIRST TRIANGLE SECOND TRIANGLE
Number of Triangles
Other Two Angles Third Side
Other Two Angles Third Side
- 1 B=42.49, C=17.51 c=86.
- 0
- 1 B=4.98, C=165.02 c=14.
- 0
- 0
- 2 B=58.05 C=76.95 c=6.89 B=121.95, C=13.05 c=1.
- 2 B=20.32, C=149.68 c=29.07 B=159.68, C=10.32 c=10.
- 2 B=87.49, C=7.51 c=45.92 B=92.51, C=2.49 c=15.
- 0
- 1 B=20.26, C=39.73 c=18.
- 2 B=105.57C=44.43 b=38.53 B=14.43C=135.57 b=9.
- 1 B=99.29C=30.71° b=193.
- 1 B=22.02, C=127.98 c=252.
- 2 B=21.90C=83.10 b=69.51 B=8.10C=96.90 b=26.
- 2 B=66.49, C=53.51 c=157.82 B=113.51, C=6.49 c=22.
- 1 B=2.87 C=27.13 c=136.
- 1 B=6.65, C=3.35 c=4.
- 0
Solutions
