The Law of Sines: Solving Non-Right Triangles 6.1

Law of Sines

Introduction to the Law of Sines

The Law of Sines is a fundamental tool in trigonometry for solving non-right triangles. It relates the sides and angles of a triangle and is especially useful when certain combinations of sides and angles are known. The Law of Sines is applicable in the following cases:

• ASA (Angle-Side-Angle): Two angles and the included side are known.

• SAA (Side-Angle-Angle): Two angles and a non-included side are known.

• SSA (Side-Side-Angle): Two sides and a non-included angle are known (the ambiguous case).

Statement of the Law of Sines

The Law of Sines states that in any triangle with sides a, b, c opposite angles A, B, C respectively:

$\displaystyle \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Usually, only two parts of the formula are used at a time to solve for an unknown side or angle.

Solving Triangles Using the Law of Sines

General Steps

1. Identify the known sides and angles.

2. Use the Law of Sines to set up a proportion involving the known values and the unknown you wish to find.

3. Solve for the unknown side or angle.

4. If necessary, use the triangle angle sum property ($A + B + C = 180^\circ$) to find missing angles.

5. Check for possible ambiguous cases (SSA), which may yield zero, one, or two solutions.

Example 1: ASA/SAA Case

Given: $A = 40^\circ$, $B = 60^\circ$, $a = 4$

• Find $C$: $C = 180^\circ - 40^\circ - 60^\circ = 80^\circ$

• Find $b$ (opposite $B$): $\displaystyle \frac{b}{\sin 60^\circ} = \frac{4}{\sin 40^\circ}$ $b = 4 \cdot \frac{\sin 60^\circ}{\sin 40^\circ} \approx 5.39$

• Find $c$ (opposite $C$): $\displaystyle \frac{c}{\sin 80^\circ} = \frac{4}{\sin 40^\circ}$ $c = 4 \cdot \frac{\sin 80^\circ}{\sin 40^\circ} \approx 6.13$

Example 2: SAA Case

Given: $A = 35^\circ$, $B = 15^\circ$, $a = 5$

• Find $C$: $C = 180^\circ - 35^\circ - 15^\circ = 130^\circ$

• Find $b$: $\displaystyle \frac{b}{\sin 15^\circ} = \frac{5}{\sin 130^\circ}$ $b = 5 \cdot \frac{\sin 15^\circ}{\sin 130^\circ} \approx 1.69$

• Find $c$: $\displaystyle \frac{c}{\sin 35^\circ} = \frac{5}{\sin 130^\circ}$ $c = 5 \cdot \frac{\sin 35^\circ}{\sin 130^\circ} \approx 2.74$

Example 3: SSA Case (Ambiguous Case)

Given: $a = 3$, $b = 2$, $A = 40^\circ$

• Find $B$: $\displaystyle \frac{2}{\sin B} = \frac{3}{\sin 40^\circ}$ $\sin B = 2 \cdot \frac{\sin 40^\circ}{3} \approx 0.4285$ $B = \arcsin(0.4285) \approx 25.37^\circ$

• Find $C$: $C = 180^\circ - 40^\circ - 25.37^\circ = 114.63^\circ$

• Find $c$: $\displaystyle \frac{c}{\sin 114.63^\circ} = \frac{3}{\sin 40^\circ}$ $c = 3 \cdot \frac{\sin 114.63^\circ}{\sin 40^\circ} \approx 4.24$

Example 4: No Solution Case

Given: $a = 1$, $b = 2$, $C = 50^\circ$

• Attempt to find $A$: $\displaystyle \frac{2}{\sin A} = \frac{1}{\sin 50^\circ}$ $\sin A = 2 \cdot \frac{\sin 50^\circ}{1} > 1$ Since $\sin A$ is greater than 1, there is no solution (the triangle cannot be constructed).

Example 5: Two Solutions (Ambiguous Case)

Given: $a = 1$, $b = 2$, $C = 25^\circ$

• Find $A$: $\displaystyle \frac{2}{\sin A} = \frac{1}{\sin 25^\circ}$ $\sin A = 2 \cdot \frac{\sin 25^\circ}{1} \approx 0.8452$ $A_1 = \arcsin(0.8452) \approx 57.7^\circ$

• Since sine is positive in the first and second quadrants, the second possible angle is $A_2 = 180^\circ - 57.7^\circ = 122.3^\circ$

• For each $A$, find $B$ and $b$:

Triangle	$A$	$B$	$b$
1	$57.7^\circ$	$180^\circ - 25^\circ - 57.7^\circ = 97.3^\circ$	$1 \cdot \frac{\sin 97.3^\circ}{\sin 25^\circ} \approx 2.35$
2	$122.3^\circ$	$180^\circ - 25^\circ - 122.3^\circ = 32.7^\circ$	$1 \cdot \frac{\sin 32.7^\circ}{\sin 25^\circ} \approx 1.28$

Note: The ambiguous case (SSA) can yield two possible triangles, one, or none, depending on the given values.

Example 6: Checking for Two Solutions

Given: $a = 4$, $b = 5$, $A = 40^\circ$

• Find $C$: $\displaystyle \frac{5}{\sin C} = \frac{4}{\sin 40^\circ}$ $\sin C = 5 \cdot \frac{\sin 40^\circ}{4} \approx 0.8035$ $C_1 = \arcsin(0.8035) \approx 53.46^\circ$

• Second possible solution: $C_2 = 180^\circ - 53.46^\circ = 126.54^\circ$

• Check if $A + C_2 < 180^\circ$; if so, two triangles are possible.

Triangle	$C$	$B$	$a$
1	$53.46^\circ$	$180^\circ - 40^\circ - 53.46^\circ = 86.54^\circ$	$4 \cdot \frac{\sin 86.54^\circ}{\sin 40^\circ} \approx 6.21$
2	$126.54^\circ$	$180^\circ - 40^\circ - 126.54^\circ = 13.46^\circ$	$4 \cdot \frac{\sin 13.46^\circ}{\sin 40^\circ} \approx 1.38$

Summary Table: Law of Sines Cases

Case	Given	Possible Solutions
ASA/SAA	2 angles, 1 side	1 solution
SSA	2 sides, non-included angle	0, 1, or 2 solutions (ambiguous case)

Key Points

• The Law of Sines is used to solve for unknown sides or angles in non-right triangles.

• It is especially useful in ASA, SAA, and SSA cases.

• The ambiguous case (SSA) may yield zero, one, or two possible triangles.

• Always check the domain of the sine function; if $\sin x > 1$ or $\sin x < -1$, there is no solution.

• Use the triangle angle sum property to find missing angles.

Practice Problems

1. Solve the triangle given $A = 30^\circ$, $B = 45^\circ$, $a = 10$.

2. Given $a = 7$, $b = 5$, $A = 40^\circ$, solve for the remaining sides and angles.

3. Given $a = 3$, $b = 4$, $C = 60^\circ$, determine if a triangle can be formed and, if so, solve it.

Additional info: The Law of Sines is foundational for solving oblique triangles and is a precursor to the Law of Cosines, which is used when two sides and the included angle (SAS) or all three sides (SSS) are known.