The Ambiguous Case

Understand ambiguous case in sine law and two-solution scenarios. Grade 11 Math 20-1.

Lesson 4.8 of Trigonometry in Math 20-1 — Alberta curriculum lessons.

What you'll learn in this lesson

• Understanding the Ambiguous Case. Discover why SSA configurations can produce zero, one, or two valid triangles. Learn how to visualize the ambiguous case using height comparisons and classify which outcome applies before solving.
• Applying the Procedure. Walk through the six-step procedure for every SSA problem. Work three parallel examples — one triangle, no triangle, and the ambiguous two-triangle case — using the same given angle and one given side to see how the third side length determines the outcome.
• Solving Full Triangles in the Ambiguous Case. When two triangles are possible, you must produce two complete solution sets. Practice solving Example 4 and an application problem about locating a missing park marker, tracking both triangles through every step.

Inside the lesson — a free preview

What Is the Ambiguous Case?

Definition — The SSA Configuration. The ambiguous case arises in trigonometry when you have an SSA (Side-Side-Angle) configuration:

Two sides of a triangle are given, AND the angle given is opposite one of the given sides — NOT between them.

Why ambiguous? Because depending on the side lengths, the same SSA information might describe zero, one, or two different triangles. Without further checking, it is not clear which case applies.

SSA vs SAS — The Key Contrast. SAS (two sides + the angle between them) always produces exactly one triangle. Use the Cosine Law.

SSA (two sides + an angle not between them) could produce 0, 1, or 2 triangles. This is the ambiguous case.

Visualizing Why This Happens

Imagine you are given angle A (acute), side b (one of the sides forming angle A), and side a (the side opposite angle A — the side that swings down from C to the base ray).

What Determines the Outcome. The third vertex B must land where side a (extending from C) meets the bottom ray. Depending on how long a is:

Too short: if a is less than the perpendicular height h, side a cannot reach the bottom ray at all — no triangle.

Just right: if a equals h exactly, it touches the bottom ray at one point (forming a right angle) — one right triangle.

Medium: if a is longer than h but shorter than b, the arc crosses the bottom ray at two points — two triangles.

Long: if a ≥ b, only one of the two crossings makes a valid triangle — one triangle.

Worked example: One Triangle (a > b)

Example 1: b = 5 cm, a = 6 cm, A = 27° — Find B. Step 1 — Compute h.

h = b · A = 5 · 27° ≈ 5 · 0.4540 ≈ 2.27

Step 2 — Compare a, h, and b.

h ≈ 2.27, a = 6, b = 5.

Since a = 6 > b = 5, we are in the case a ≥ b — only one triangle exists.

Step 3 — Apply the Sine Law to find B.

( B)/(b) = ( A)/(a)

( B)/(5) = ( 27°)/(6)

B = (5 · 27°)/(6) ≈ (5 · 0.4540)/(6) ≈ (2.270)/(6) ≈ 0.3783

Step 4 — Take the inverse sine.

B = ^(-1)(0.3783) ≈ 22.2° ≈ 22°

Step 5 — Check the obtuse case (to confirm rejection).

If B were 180° - 22° = 158°, then A + B = 27° + 158° = 185° > 180°. Reject.

Answer: B ≈ 22° — only one possibility.

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