The Law of Cosines: Solving Non-Right Triangles 6.2

Law of Cosines

Introduction to the Law of Cosines

The Law of Cosines is a fundamental formula in trigonometry used to solve for unknown sides or angles in any triangle, not just right triangles. It is especially useful in cases where the Law of Sines cannot be directly applied, specifically when given two sides and the included angle (SAS) or all three sides (SSS) of a triangle.

• SAS (Side-Angle-Side): Two sides and the included angle are known.

• SSS (Side-Side-Side): All three sides are known.

Formulas for the Law of Cosines

The Law of Cosines can be written in three equivalent forms, depending on which side or angle you are solving for:

• $c^2 = a^2 + b^2 - 2ab \cos C$

• $a^2 = b^2 + c^2 - 2bc \cos A$

• $b^2 = a^2 + c^2 - 2ac \cos B$

Where a, b, and c are the lengths of the sides of the triangle, and A, B, and C are the measures of the angles opposite those sides, respectively.

Solving Triangles Using the Law of Cosines

To solve a triangle means to find all unknown sides and angles. The Law of Cosines is applied as follows:

• Use the appropriate formula based on the known values.

• After finding one side or angle, use the Law of Sines or the Law of Cosines again to find the remaining parts.

• Check for possible ambiguous cases, especially when using the Law of Sines.

Example 1: SAS Case

Given: $a = 2$, $b = 3$, $C = 60^\circ$

• Find side $c$ using $c^2 = a^2 + b^2 - 2ab \cos C$:

• $c^2 = 2^2 + 3^2 - 2 \times 2 \times 3 \cos 60^\circ$

• $c^2 = 4 + 9 - 12 \times 0.5 = 13 - 6 = 7$

• $c = \sqrt{7}$

• Find angle $A$ using the Law of Sines:

• $\dfrac{\sin A}{a} = \dfrac{\sin C}{c}$

• $\sin A = \dfrac{a \sin C}{c} = \dfrac{2 \sin 60^\circ}{\sqrt{7}}$

• $\sin A \approx 0.6547 \implies A \approx 40.89^\circ$

• Find angle $B$:

• $B = 180^\circ - C - A = 180^\circ - 60^\circ - 40.89^\circ = 79.11^\circ$

Note: Only one solution is possible when all three sides are known.

Example 2: SAS Case with Obtuse Angle

Given: $a = 3$, $c = 2$, $B = 91^\circ$

• Find side $b$ using $b^2 = a^2 + c^2 - 2ac \cos B$:

• $b^2 = 3^2 + 2^2 - 2 \times 3 \times 2 \cos 91^\circ$

• $\cos 91^\circ \approx -0.0175$

• $b^2 = 9 + 4 - 12 \times (-0.0175) = 13 + 0.21 = 13.21$

• $b \approx 3.63$

• Find angle $A$ using the Law of Sines:

• $\dfrac{\sin A}{a} = \dfrac{\sin B}{b}$

• $\sin A = \dfrac{a \sin B}{b} = \dfrac{3 \sin 91^\circ}{3.63}$

• $\sin A \approx 0.8263 \implies A \approx 55.72^\circ$

• Find angle $C$:

• $C = 180^\circ - B - A = 180^\circ - 91^\circ - 55.72^\circ = 33.28^\circ$

Example 3: SSS Case

Given: $a = 4$, $b = 3$, $c = 6$

• Find angle $A$ using $a^2 = b^2 + c^2 - 2bc \cos A$:

• $4^2 = 3^2 + 6^2 - 2 \times 3 \times 6 \cos A$

• $16 = 9 + 36 - 36 \cos A$

• $16 = 45 - 36 \cos A$

• $36 \cos A = 45 - 16 = 29$

• $\cos A = 29/36 \approx 0.8055$

• $A \approx 36.34^\circ$

• Find angle $C$ using Law of Sines or Law of Cosines:

• Using Law of Sines: $\sin C = \dfrac{c \sin A}{a} = \dfrac{6 \sin 36.34^\circ}{4}$

• $\sin C \approx 0.8889 \implies C \approx 62.73^\circ$

• However, since $C$ should be obtuse (from the diagram), $C = 180^\circ - 62.73^\circ = 117.27^\circ$

• Find angle $B$:

• $B = 180^\circ - A - C = 180^\circ - 36.34^\circ - 117.27^\circ = 26.38^\circ$

Additional info: When all three sides are given, using the Law of Cosines for each angle avoids ambiguity.

Example 4: Isosceles Triangle (SSS)

Given: $a = 3$, $b = 3$, $c = 2$

• Since $a = b$, triangle is isosceles, so $A = B$.

• Find angle $A$ using $a^2 = b^2 + c^2 - 2bc \cos A$:

• $3^2 = 3^2 + 2^2 - 2 \times 3 \times 2 \cos A$

• $9 = 9 + 4 - 12 \cos A$

• $9 = 13 - 12 \cos A$

• $12 \cos A = 13 - 9 = 4$

• $\cos A = 4/12 = 0.3333$

• $A \approx 70.53^\circ$

• So $B \approx 70.53^\circ$

• Find angle $C$:

• $C = 180^\circ - 70.53^\circ - 70.53^\circ = 38.94^\circ$

Example 5: Application – Navigation Problem

Scenario: A plane flies due north from Ft. Myers to Sarasota (150 miles), then at a bearing of $50^\circ$ east of north to Orlando (100 miles).

• Let $A$ = Ft. Myers, $B$ = Sarasota, $C$ = Orlando.

• Angle at $B$ inside the triangle: $B = 180^\circ - 50^\circ = 130^\circ$

• Find distance from Ft. Myers to Orlando ($b$):

• $b^2 = 100^2 + 150^2 - 2 \times 100 \times 150 \cos 130^\circ$

• $\cos 130^\circ \approx -0.6428$

• $b^2 = 10000 + 22500 - 30000 \times (-0.6428) = 32500 + 19284 = 51784$

• $b = \sqrt{51784} \approx 227.56$ miles

• Find bearing from Ft. Myers to Orlando (angle $A$):

• Use $a = 100$, $b = 227.56$, $c = 150$

• $\cos A = \dfrac{100^2 + 227.56^2 - 150^2}{2 \times 100 \times 227.56}$

• Alternatively, as shown in the notes, $\cos A \approx 0.9416$

• $A \approx 19.67^\circ$

• Answer: The pilot should fly on a bearing of $N 19.67^\circ E$ from Ft. Myers to Orlando.

Summary Table: Law of Cosines Cases

Case	Known	Unknown	First Step
SAS	Two sides and included angle	Third side	Use Law of Cosines to find unknown side
SSS	All three sides	All angles	Use Law of Cosines to find one angle, then others

Key Points

• The Law of Cosines generalizes the Pythagorean Theorem for all triangles.

• It is essential for solving triangles when the Law of Sines is not applicable.

• Always check for possible ambiguous cases when using the Law of Sines after the Law of Cosines.

• For SSS cases, using the Law of Cosines for each angle avoids ambiguity.