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The Ambiguous SSA Case

A free Trigonometry lesson from the “Applications of Trigonometry” unit, with a worked example and practice problems including step-by-step solutions.

In the SSA case you know two sides and an angle not between them, so the triangle may not be unique. Compute the height h = b*sin(A) from the known angle's vertex and compare it to side a: if a < h there are 0 triangles, if a = h or a >= b there is exactly 1, and if h < a < b there are 2. When two triangles exist, the Law of Sines gives sin(B), and both an acute B and its obtuse supplement 180 - B are valid second angles.

What you'll learn

Why it matters: Surveyors and navigators who measure one angle and two distances can hit this ambiguity, where the same readings describe two genuinely different triangles, so they must test for the second solution before trusting a single position.

Worked example

Problem. In triangle ABC, A = 35 degrees, a = 9, and b = 12. Determine how many triangles are possible, then find both possible measures of angle B to the nearest tenth.

  1. Height h = b*sin(A) = 12*sin(35) = 6.9; since h < a < b (6.9 < 9 < 12), two triangles exist.
  2. Law of Sines: sin(B) = b*sin(A)/a = 12*sin(35)/9 = 0.7648.
  3. B1 = sin^-1(0.7648) = 49.9 degrees, and B2 = 180 - 49.9 = 130.1 degrees.

Answer: 2 triangles; B = 49.9 degrees or B = 130.1 degrees

Practice problems

1. In an SSA setup with A = 30 degrees and b = 10, compute the height h = b*sin(A).

Show solution
  1. Warm-up: First identify exactly what the question is asking: In an SSA setup with A = 30 degrees and b = 10, compute the height h = b*sin(A).
  2. Use inverse operations to isolate the unknown, and keep both sides balanced at every step.
  3. The height is h = b*sin(A).
  4. h = 10*sin(30) = 10*(1/2).
  5. h = 5.
  6. Check the result by substituting or estimating: the response should match 5 and make sense in the original problem.

Answer: 5

2. In the SSA case, side a is the side opposite the known angle A. The height used for the decision test is measured from which vertex?

Choices: h = b*sin(A), the altitude from vertex C to side c · h = a*sin(B), the altitude from vertex A · h = c*sin(A), the altitude from vertex B · h = b*cos(A), a horizontal distance

Show solution
  1. The known angle is A, with given sides a and b.
  2. Drop the altitude from C; its length is b*sin(A).
  3. Side a is then compared to this height h.

Answer: h = b*sin(A), the altitude from vertex C to side c

3. A = 30 degrees, a = 8, b = 10. Compute h = b*sin(A) and compare it to a: how many triangles are possible?

Show solution
  1. Warm-up: First identify exactly what the question is asking: A = 30 degrees, a = 8, b = 10. Compute h = b*sin(A) and compare it to a: how many triangles are possible?
  2. Use the relevant geometric relationship first, then set up an equation from the angle measures or side relationships.
  3. h = 10*sin(30) = 5.
  4. Since h < a < b (5 < 8 < 10), two triangles exist.
  5. Answer: 2.
  6. Check the result by substituting or estimating: the response should match 2 and make sense in the original problem.

Answer: 2

4. A = 40 degrees, a = 6, b = 10. Find h = b*sin(A) to the nearest tenth, then state the number of triangles.

Show solution
  1. Core Practice: First identify exactly what the question is asking: A = 40 degrees, a = 6, b = 10. Find h = b*sin(A) to the nearest tenth, then state the number of triangles.
  2. Use the relevant geometric relationship first, then set up an equation from the angle measures or side relationships.
  3. h = 10*sin(40) = 6.4.
  4. Since a = 6 < h = 6.4, no triangle can be drawn.
  5. Number of triangles: 0.
  6. Check the result by substituting or estimating: the response should match 0 and make sense in the original problem.

Answer: 0

5. A = 35 degrees, a = 12, b = 9. How many triangles are possible?

Show solution
  1. Core Practice: First identify exactly what the question is asking: A = 35 degrees, a = 12, b = 9. How many triangles are possible?
  2. Use the relevant geometric relationship first, then set up an equation from the angle measures or side relationships.
  3. Here a = 12 is greater than b = 9.
  4. When a >= b with A acute, exactly one triangle exists.
  5. Number of triangles: 1.
  6. Check the result by substituting or estimating: the response should match 1 and make sense in the original problem.

Answer: 1

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