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TR3 · Solve triangles with the sine and cosine laws, including SSA ambiguity

Learn to solve triangles with the sine and cosine laws, including ssa ambiguity through clear examples and targeted practice.

Alberta Grade 11 Mathematics

Trigonometry

Why one set of measurements can give two triangles

The sine law is useful when a known angle is paired with its opposite side. But side–side–angle (SSA) information can make two different triangles. A sketch alone is not enough: compare the side opposite the given angle with the altitude, then check the angle sum for every candidate. This lesson uses degree measure, as required in Mathematics 20-1.

What you will learn

  • Recognize when SSA data can produce two triangles.
  • Calculate both valid triangles using the sine law.
  • Reject any candidate whose angles cannot form a triangle.

Match each angle to its opposite side

Let a be opposite angle A and b opposite angle B. The sine law equates side length divided by the sine of its opposite angle. Match the labels carefully before substituting numbers.
The inverse sine gives one angle between 0° and 90°, but the supplementary angle can have the same sine. Do not assume the calculator’s first answer is the only possible angle.
asin⁡A\frac{a}{\sin A}=bsin⁡B\frac{b}{\sin B}=csin⁡C\frac{c}{\sin C}

Test each possible triangle

For an acute given angle A, two triangles are possible when the opposite side a is longer than the altitude bsin⁡Ab\sin A but shorter than b. The diagram is calculated from the given lengths rather than drawn freehand.
For each candidate B, find C from A+B+C=180∘A+B+C=180^\circ. A candidate is valid only when C is positive. Then use the sine law to find the remaining side c. Small differences are expected when you round angles.
bsin⁡A<a<bb\sin A<a<b

Worked example

Two triangles from the same data

In triangle ABC, A=30∘A=30^\circ, a=7a=7 cm, and b=10b=10 cm. Find both possible values of B, C, and c.
Two possible SSA triangles
Two possible SSA trianglesBoth triangles use A = 30°, a = 7 cm, and b = 10 cm. The computed third side differs between the two configurations.ACBb = 10 cma = 7 cmc = 13.56 cm30°Solution 1ACBb = 10 cma = 7 cmc = 3.76 cm30°Solution 2

Both triangles use A = 30°, a = 7 cm, and b = 10 cm. The computed third side differs between the two configurations.

  1. Check for ambiguity
    The altitude is 10sin⁡30∘=510\sin30^\circ=5 cm. Since 5 < 7 < 10, two triangles are possible.
    10sin⁡30°=5<7<1010\sin30°=5<7<10
  2. Find the two B angles
    The sine law gives sin⁡B=5/7\sin B=5/7. The acute angle is about 45.58° and its supplement is about 134.42°.
    B1≈45.58°,B2≈134.42°B_1\approx45.58°, B_2\approx134.42°
  3. Complete each triangle
    Subtract A and B from 180°. Then use c/sin⁡C=7/sin⁡30∘=14c/\sin C=7/\sin30^\circ=14.
    C_1\approx104.42°,\ c_1\approx13.56 cm\mathrm{cm}; C_2\approx15.58°,\ c_2\approx3.76 cm\mathrm{cm}
Answer: The two triangles have (B,C,c)≈(45.58∘,104.42∘,13.56 cm)(B,C,c)\approx(45.58^\circ,104.42^\circ,13.56\,\mathrm{cm}) or (134.42∘,15.58∘,3.76 cm)(134.42^\circ,15.58^\circ,3.76\,\mathrm{cm}).
Check: Each angle triple sums to 180° and each computed c is positive; both satisfy the given side and angle data, up to rounding.

Common mistakes and how to avoid them

Keeping only the calculator’s acute angle for B.
Correction: Also test its supplement because both angles have the same sine.
Accepting a triangle without checking the angle sum.
Correction: Calculate C and require it to be greater than zero.

Lesson summary

  • SSA can have zero, one, or two triangles.
  • For this data, the altitude test predicts two.
  • Solve and verify each angle set separately.

Check your understanding

Question 1

If sin⁡B=1/2\sin B=1/2 in a valid SSA triangle, which B values should you test?
  1. 30∘30^\circ and 150∘150^\circ
  2. Only 30∘30^\circ
  3. 30∘30^\circ and 60∘60^\circ
Show answer and explanation
30∘30^\circ and 150∘150^\circ
Supplementary angles have the same sine; the total angle sum decides which candidates remain valid.

Key terms

SSA
Two sides and an angle opposite one of them are given.
Ambiguous case
A set of SSA measurements that forms more than one triangle.

Continue through Math 20-1

View the complete Math 20-1 Alberta Grade 11 Mathematics curriculum and lessons

About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Alberta Grade 11 Mathematics (Math 20-1), expectation TR3. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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