7.2: The Ambiguous Case | Precalculus | Code Scholars Tutoring

How it works

With SSA data and an acute known angle, a side can reach a ray at zero, one, or two points. After an inverse-sine calculation, test the supplementary angle and verify the angle sum. A sine value greater than one means no triangle.

\sin B=\frac{b\sin A}{a},\quad B_2=180^\circ-B_1

Worked examples

Example 1

Given $\displaystyle A=30^\circ$ , $\displaystyle a=5$ , $\displaystyle b=8$ , determine the number of triangles.

$\displaystyle \sin B=0.8$ . Both $\displaystyle B\approx53.13^\circ$ and $\displaystyle 126.87^\circ$ leave a positive third angle, so there are two triangles.

Example 2

Given $\displaystyle A=30^\circ$ , $\displaystyle a=3$ , $\displaystyle b=8$ , determine the number of triangles.

$\displaystyle \sin B=8(1/2)/3=4/3>1$ , so there is no triangle.

Problems

Problem 7.2.1

For $\displaystyle A=30^\circ$ , $\displaystyle a=5$ , $\displaystyle b=8$ , find the two possible values of C to two decimals.

Show worked solution

The possible B angles are $\displaystyle 53.13^\circ$ and $\displaystyle 126.87^\circ$ . Subtract each and A from $\displaystyle 180^\circ$ to obtain $\displaystyle 96.87^\circ$ and $\displaystyle 23.13^\circ$ .

Problem 7.2.2

For $\displaystyle A=30^\circ$ , $\displaystyle a=4$ , $\displaystyle b=8$ , how many triangles exist?

Show worked solution

$\displaystyle \sin B=1$ , so $\displaystyle B=90^\circ$ and $\displaystyle C=60^\circ$ . Exactly one triangle exists.

Problem 7.2.3

For $\displaystyle A=30^\circ$ , $\displaystyle a=10$ , $\displaystyle b=8$ , explain why only one inverse-sine candidate works.

Show worked solution

$\displaystyle B_1\approx23.58^\circ$ works; $\displaystyle B_2\approx156.42^\circ$ gives $\displaystyle A+B_2>180^\circ$ and must be rejected.

Lesson Problems PDF Solutions PDF Unit MCQ

How it works

Worked examples

Example 1

Example 2

Problems

Problem 7.2.1

Problem 7.2.2

Problem 7.2.3

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