Grade by Grade
D. Trigonometric Functions · Trigonometric Ratios, Identities and Laws

The Ambiguous Case of the Sine Law

D1.6About 75 minutes
Learning goals
  • I can explain why two sides and a non-included angle (SSA) may give no triangle, one triangle or two triangles.
  • I can compare a with h = b sin A and with b to decide how many triangles are possible.
  • I can solve both triangles when the ambiguous case gives two solutions.
  • I can interpret two-solution answers in real situations.

Words to know

SSA
Two sides and an angle that is not between them (a non-included angle).
ambiguous case
The SSA situation, where the given information may produce zero, one or two different triangles.
height h
In the ambiguous case with ∠A, side b and opposite side a: h = b sin A, the shortest distance from C to the side opposite it.
supplementary angles
Two angles that add to 180°. They have the same sine.

Same instructions, different triangles

Suppose a friend says: "Draw a triangle with ∠A = 30°, AC = 10 cm, and BC = 7 cm." You draw yours, a friend draws theirs, and you get two different triangles, both correct! This is the ambiguous case. It happens only when you know two sides and an angle that is not between them (SSA). In Grade 10 every triangle was acute, so this never came up. Now it can.

Why it happens

Draw ∠A and side b = AC. Side a must reach from C down to the base line (the other arm of ∠A). Think of side a as a gate swinging on a hinge at C. The shortest it can be and still reach the base is the height h = b sin A.

  • If a is shorter than h, it can't reach: no triangle.
  • If a equals h, it just touches: one right triangle.
  • If a is between h and b, it can reach the base in two places: two triangles.
  • If a is at least b, it reaches the base once on the far side (the other crossing is behind A, not on the arm of the angle): one triangle.
30°ACB₂B₁b = 10h = 5a = 7a = 7Swing side aaround C.It meets thebase twice,so there aretwo triangles.
∠A = 30°, b = 10 and a = 7. Since h = 10 sin 30° = 5 and 5 < 7 < 10, side a can land at B₁ or at B₂. Both triangles have the same given parts.
a < hno trianglea = hone right triangleh < a < btwo trianglesa ≥ bone triangle
The four possibilities when ∠A is acute. The dashed curve shows every point that side a can reach from C.

Checklist for SSA (given ∠A, b and a)

If ∠A is acute, find h = b sin A, then:

  • a < h: no triangle
  • a = h: one (right) triangle
  • h < a < b: two triangles
  • a ≥ b: one triangle

If ∠A is right or obtuse: a ≤ b gives no triangle; a > b gives one triangle.

In the two-triangle case, the sine law gives an acute angle β for ∠B. The second triangle uses 180° − β.

Worked example

Example 1: No triangle

∠A = 30°, b = 10 cm, a = 4 cm. How many triangles are possible?

  1. h = 10 sin 30° = 5 cm.
  2. a = 4 < 5 = h, so side a is too short to reach.
  3. (The sine law agrees: sin B = 10 sin 30° ÷ 4 = 1.25, which is impossible.)

No triangle.

Worked example

Example 2: One right triangle

∠A = 30°, b = 10 cm, a = 5 cm. How many triangles are possible?

  1. h = 5 cm, and a = h.
  2. sin B = 10 sin 30° ÷ 5 = 1, so ∠B = 90°.

One right triangle, with ∠B = 90°, ∠C = 60° and c = 10 cos 30° = 5√3 ≈ 8.7 cm.

Worked example

Example 3: Two triangles

∠A = 30°, b = 10 cm, a = 7 cm. Solve all possible triangles.

  1. h = 5. Since 5 < 7 < 10, there are two triangles.
  2. sin B = 10 sin 30° ÷ 7 = 5/7 ≈ 0.7143.
  3. Triangle 1: ∠B₁ = sin⁻¹(0.7143) ≈ 45.6°. ∠C₁ ≈ 180° − 30° − 45.6° = 104.4°. c₁ = 7 sin 104.4° ÷ sin 30° ≈ 13.6 cm.
  4. Triangle 2: ∠B₂ ≈ 180° − 45.6° = 134.4°. ∠C₂ ≈ 180° − 30° − 134.4° = 15.6°. c₂ = 7 sin 15.6° ÷ sin 30° ≈ 3.8 cm.

Triangle 1: ∠B ≈ 45.6°, ∠C ≈ 104.4°, c ≈ 13.6 cm. Triangle 2: ∠B ≈ 134.4°, ∠C ≈ 15.6°, c ≈ 3.8 cm.

Worked example

Example 4: One triangle (a ≥ b)

∠A = 30°, b = 10 cm, a = 12 cm. Solve the triangle.

  1. a = 12 ≥ b = 10, so there is only one triangle.
  2. sin B = 10 sin 30° ÷ 12 ≈ 0.4167, so ∠B ≈ 24.6°.
  3. Why not 180° − 24.6° = 155.4°? Then ∠A + ∠B = 185.4° > 180°. Impossible.
  4. ∠C ≈ 180° − 30° − 24.6° = 125.4°, and c = 12 sin 125.4° ÷ sin 30° ≈ 19.6 cm.

∠B ≈ 24.6°, ∠C ≈ 125.4°, c ≈ 19.6 cm

Worked example

Example 5: Driving into radio range

A straight highway leaves town A. A radio transmitter T is 20 km from A, and the angle between the highway and line AT is 25°. The transmitter's signal reaches 12 km. Along which stretch of the highway can a driver hear the station?

  1. A point P on the highway gets the signal if TP ≤ 12 km. The edges of that stretch are where TP = 12 exactly: ∠A = 25°, AT = 20, TP = 12 (SSA).
  2. h = 20 sin 25° ≈ 8.45. Since 8.45 < 12 < 20, there are two points.
  3. sin P = 20 sin 25° ÷ 12 ≈ 0.7044, so ∠P ≈ 44.8° or ∠P ≈ 135.2°.
  4. If ∠P ≈ 44.8°: ∠T ≈ 110.2°, so AP = 12 sin 110.2° ÷ sin 25° ≈ 26.6 km.
  5. If ∠P ≈ 135.2°: ∠T ≈ 19.8°, so AP = 12 sin 19.8° ÷ sin 25° ≈ 9.6 km.

The driver hears the station from about 9.6 km to 26.6 km from town A, a stretch of about 17 km. Here both triangles have a real meaning.

Just for fun

The ambiguous case is the only part of this course where the honest answer to "How many triangles?" can be "It depends." Mathematicians call that a case analysis, which sounds much more official.

Hands-on activity

Swing the gate yourself

You need: paper, a ruler, a protractor, a pencil, and a strip of thin cardboard or paper (or a piece of string) with a pushpin or your fingertip as a pivot.

  1. Draw a horizontal line about 20 cm long. Near its left end mark A. Draw a 30° angle at A and make AC = 10 cm.
  2. Cut (or mark) a strip exactly 7 cm long. Hold one end at C and swing the other end down to the base line. Mark every point where it touches the line. (Use scissors carefully, or just mark the length on the strip.)
  3. Repeat with strips of 4 cm, 5 cm and 12 cm.
  4. For each strip, record how many triangles you could make.
  5. Measure ∠B in each triangle you made and compare with the calculations in Examples 2 to 4.

What to look for: 4 cm never reaches the line (no triangle); 5 cm just touches it at a right angle (one triangle); 7 cm reaches it twice, about 4.9 cm on either side of the foot of the height (√(7² − 5²) = √24) (two triangles, with ∠B ≈ 46° and 134°); 12 cm reaches it only on the far side (one triangle).

Your turn

1

∠A = 40°, b = 12 cm, a = 6 cm. How many triangles?

2

∠A = 40°, b = 12 cm, a = 10 cm. Solve all possible triangles.

3

∠A = 40°, b = 12 cm, a = 15 cm. Solve the triangle.

4

∠A = 120°, b = 8 cm, a = 6 cm. How many triangles?

5

∠A = 120°, b = 8 cm, a = 11 cm. Solve the triangle.

6

∠A = 30°, b = 14 cm, a = 7 cm. How many triangles? Describe it.

More practice: warm-up

1

Write the formula for the height h in the ambiguous case.

2

∠A = 30° and b = 20. Find h.

3

∠A = 30°, b = 20, a = 8. How many triangles?

4

∠A = 30°, b = 20, a = 25. How many triangles?

5

∠A = 30°, b = 20, a = 15. How many triangles?

More practice: on your own

1

Solve both triangles: ∠A = 35°, b = 15 cm, a = 10 cm.

2

In △PQR, ∠Q = 28°, q = 9 cm and r = 16 cm. Find all possible measures of ∠R.

3

Explain why ∠A acute and a ≥ b gives only one triangle.

4

Two straight fences meet at a 32° angle. One is 20 m long. A third straight fence, 13 m long, joins the end of the 20 m fence to the second fence, closing a triangular garden. How long could the second fence be?

5

∠A = 60°, b = 10 cm, a = 9 cm. How many triangles? Find each possible ∠B.

Challenge

1

Redo Example 5 for a weaker transmitter with a range of 10 km. Along which stretch can the station be heard?

2

∠A = 50° and b = 10. For what values of a are there exactly two triangles?

3

Use the cosine law to solve Example 3 for c instead: 7² = 10² + c² − 2(10)(c) cos 30°. Explain how this quadratic shows the ambiguous case.

Check your understanding

Quiz

1.

The ambiguous case occurs when you are given:

2.

∠A = 30°, b = 10, a = 4. How many triangles are possible?

3.

∠A = 30°, b = 10, a = 7. How many triangles are possible?

4.

In the two-triangle case the sine law gives ∠B ≈ 45.6°. What is ∠B in the second triangle?

5.

∠A = 120°, b = 8 and a = 6. How many triangles?

6.

∠A is acute and a ≥ b. How many triangles?

For parents and teachers

Big idea (D1.6, ambiguous case): when the given parts are two sides and an angle not between them (SSA), there may be no triangle, one triangle, or two different triangles. Comparing side a with the height h = b sin A (and with b) tells which.

How to help: The "swinging gate" activity is the best way to understand this. If your teen is unsure, ask them to draw the angle and side b to scale and physically swing a strip of length a.

Common difficulties:

  • Using the sine law, getting one angle, and stopping without checking for a second triangle.
  • Forgetting that the second angle is 180° − β, not 90° − β or 360° − β.
  • Reporting a "second triangle" whose angles add to more than 180°. Always check the angle sum.