Law of Sines and the Ambiguous Case

Develop corresponding side-angle pairs, ssa feasibility and ambiguity.

Lesson 1 of 42 parts Practice Flashcards

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Corresponding Side-Angle Pairs

Objective: Explain corresponding side-angle pairs, apply the rule, and justify relevant restrictions.

In any triangle, a/sin A=b/sin B=c/sin C, with each lowercase side opposite its matching angle. Use the law when a known opposite side-angle pair is available. For ASA or AAS data, find the third angle from A+B+C=π, then solve missing sides. The equation does not permit mixing a side with a nonopposite angle.

Formulas

$a/sin A=b/sin B=c/sin C$

Worked examples

Try itIn a triangle, A=30°, B=90°, and side a=7 opposite A. Find side b opposite B.Show answer

Step 1: Use a/sin A=b/sin B. Step 2: Substitute 7/(1/2)=b/1. Step 3: Therefore b=14.

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Corresponding Side-Angle Pairs

Objective: Explain corresponding side-angle pairs, apply the rule, and justify relevant restrictions.

Formulas

$a/sin A=b/sin B=c/sin C$

Worked examples

Try itIn a triangle, A=30°, B=90°, and side a=7 opposite A. Find side b opposite B.Show answer

Step 1: Use a/sin A=b/sin B. Step 2: Substitute 7/(1/2)=b/1. Step 3: Therefore b=14.

SSA Feasibility and Ambiguity

Objective: Explain ssa feasibility and ambiguity, apply the rule, and justify relevant restrictions.

Given angle A and sides a and b in an SSA configuration, calculate sin B=b sin A/a. A required value greater than one rules out a triangle. For an admissible sine, consider both B and π−B, retaining only choices for which A+B<π. Depending on the data, zero, one, or two triangles can result; principal arcsine alone does not settle the count.

Formulas

$sin B=b sin A/a; A+B<π$

Worked examples

Try itFor SSA data with acute A=30°, side a=7, and side b=21, find the value the sine law would require for sin B.Show answer

Step 1: Use sin B=b sin A/a. Step 2: Substitute (21)(1/2)/7. Step 3: The required value is 3/2>1, so no triangle exists.

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