MathematicsChapter 8: Application of Trigonometry
Ambiguous Case Ssa (school)
For an acute given angle A, let a be its opposite side, b the other known side, and h = b sin A.
Learning objectives
- Explain ambiguous case ssa (school) using the correct definitions, conditions, and notation.
- Apply h = b sin A, a vs h vs b correctly in routine and exam-style questions.
- Show each important step, simplify the result, and check that it satisfies the original conditions.
Definition
For an acute given angle A, let a be its opposite side, b the other known side, and h = b sin A. If a < h, no triangle exists. If a = h, there is one right triangle. If h < a < b, there are two triangles. If a >= b, there is one triangle.
Mathematical form
h = b sin A
h = b sin A. If a
a = h
h = b sin A a
Where / symbols
- AMBIGUOUS CASE SSA (SCHOOL)
- For an acute given angle A, let a be its opposite side, b the other known side, and h = b sin A. If a < h, no triangle exists. If a = h, there is one right triangle. If h < a < b, there are two triangles. If a >= b, there is one triangle.
- NO TRIANGLE / TWO
- If a < h, side a cannot reach the base, so no triangle exists. If a = h, it meets the base once at 90 degrees. If h < a < b, it can meet the base in two places, giving two triangles.
- ONE TRIANGLE
- If a >= b for acute A, there is one triangle. The complete count is therefore 0 when a<h, 1 when a=h, 2 when h<a<b, and 1 when a>=b.
Relationship
- h = b sin A
Overview
Compare a with the height:
No Triangle / Two
If a < h, side a cannot reach the base, so no triangle exists. If a = h, it meets the base once at 90 degrees. If h < a < b, it can meet the base in two places, giving two triangles.
Examples h = b sin A · h < a < b => two · Sketch mentally
One Triangle
If a >= b for acute A, there is one triangle. The complete count is therefore 0 when a<h, 1 when a=h, 2 when h<a<b, and 1 when a>=b.
Examples a = h right triangle · Acute A assumed here · Do not force two always
WHAT — AMBIGUOUS CASE SSA (SCHOOL)
For an acute given angle A, let a be its opposite side, b the other known side, and h = b sin A. If a < h, no triangle exists. If a = h, there is one right triangle. If h < a < b, there are two triangles. If a >= b, there is one triangle. The symbols and conditions are part of the mathematics, so include them when stating the rule.
WHY — Why this topic matters
Ambiguous Case Ssa (School) matters because trigonometry links angles and lengths. It lets us solve triangles and practical height, distance, bearing, and area problems without measuring every side directly.
HOW — How to solve and explain it
Start by identifying what is given and what must be found. Work through NO TRIANGLE / TWO, then ONE TRIANGLE. Worked: acute A, a < b sin A. How many triangles? h = b sin A a < h side a cannot reach the other side number of triangles = 0 Finish by checking signs, restrictions, units, and whether the answer matches the question.
Worked example
- acute A, a < b sin A. How many triangles?
- h = b sin A
- a < h
- side a cannot reach the other side
- number of triangles = 0
In simple words
Do not leave out the boundary a = h: it gives exactly one right triangle. The two-triangle case uses strict inequalities h < a < b.
Key points
- h = b sin A
- a vs h vs b
- a < h => none
- a > b => one
Exam preparation
Exam Focus
Quick revision
Revision Zone
Review the key facts, formulas and checks before you move on.Ambiguous Case Ssa (school)
Sticky Notes · Quick Revision
Core Idea
- For an acute given angle A, let a be its opposite side, b the other known side, and h = b sin A. If a < h, no triangle exists. If a = h, there is one right triangle. If h < a < b, there are two triangles. If a >= b, there is one triangle.
NO TRIANGLE / TWO
- If a < h, side a cannot reach the base, so no triangle exists. If a = h, it meets the base once at 90 degrees. If h < a < b, it can meet the base in two places, giving two triangles.
ONE TRIANGLE
- If a >= b for acute A, there is one triangle. The complete count is therefore 0 when a<h, 1 when a=h, 2 when h<a<b, and 1 when a>=b.
Worked Method
- Worked: acute A, a < b sin A. How many triangles?
- h = b sin A
- a < h
- side a cannot reach the other side
- number of triangles = 0
Common Misconception
- Do not leave out the boundary a = h: it gives exactly one right triangle. The two-triangle case uses strict inequalities h < a < b.
FBISE Exam Check
- Do not leave out the boundary a = h: it gives exactly one right triangle. The two-triangle case uses strict inequalities h < a < b.
60-second revision
- For an acute given angle A, let a be its opposite side, b the other known side, and h = b sin A.
- NO TRIANGLE / TWO: If a < h, side a cannot reach the base, so no triangle exists.
- ONE TRIANGLE: If a >= b for acute A, there is one triangle.
- Worked-method reminder: number of triangles = 0
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