Law of Cosines or the Generalized Pythagorean Theorem
Law of Cosines or the Generalized Pythagorean Theorem in a Scalene Triangle. Proof
The Law of Cosines. Formula and its proof
The Law of Cosines, or the generalized Pythagorean theorem
The Law of Cosines, or the generalized Pythagorean theorem states that in any triangle △ABC we have these three relationships between the lengths of the sides a, b and c and the cosine of the angles ∠A, ∠B and ∠C:
- a2 = b2 + c2 - 2 × b × c × cos(∠A)
- b2 = a2 + c2 - 2 × a × c × cos(∠B)
- c2 = a2 + b2 - 2 × a × b × cos(∠C)
- Where:
- a = the side of the triangle opposite angle ∠A
- b = the side of the triangle opposite angle ∠B
- c = the side of the triangle opposite angle ∠C
It is also called the generalized Pythagorean theorem because it works for all triangles. When the angle ∠A, ∠B, or ∠C is a right angle cos(90°) = 0, making the extra part disappear and leaving the classic - The Pythagorean theorem:
- a2 = b2 + c2
- b2 = a2 + c2
- c2 = a2 + b2.
Proof of the Law of Cosines
- In any triangle △ABC, which is not right-angled, with side lengths BC = a, AC = b and AB = c, we draw from the vertex A the perpendicular to the side BC, which intersects BC in point D. We note the lengths of the sides AD = ha, BD = a1 and CD = a2, where a1 + a2 = a.
Proof of the Law of Cosine
In the right triangle △ABD we have the following trigonometric relations:
- sin(∠B) = ha/c => ha = c × sin(∠B)
- cos(∠B) = a1/c => a1 = c × cos(∠B)
In the right triangle △ADC we apply the Pythagorean Theorem:
- b2 = ha2 + a22
- b2 = ha2 + (a - a1)2
Substitute for ha and a1 in the formula above:
- b2 = c2 × sin2(∠B) + (a - c × cos(∠B))2
- b2 = c2 × sin2(∠B) + a2 - 2 × a × c × cos(∠B) + c2 × cos2(∠B)
- b2 = c2 × (sin2(∠B) + cos2(∠B)) + a2 - 2 × a × c × cos(∠B)
- b2 = c2 × (ha2/c2 + a12/c2) + a2 - 2 × a × c × cos(∠B)
- b2 = c2 × (ha2 + a12)/c2 + a2 - 2 × a × c × cos(∠B)
In the right triangle △ABD apply Pythagora's Theorem:
- ha2 + a12 = c2
- (ha2 + a12)/c2 = 1
Substitute for (ha2 + a12)/c2 in the last formula of b2:
b2 = c2 + a2 - 2 × a × c × cos(∠B)
By drawing perpendiculars hb and hc to the other two sides of the triangle we could also proof, in the same way, the other two relationships of the Law of Cosines.
When to use it
- The Law of Cosines could be used to:
- 1. Find an unknown length of a side of a triangle if we know the other two sides' lengths and the measure of the opposite angle (Side-Angle-Side SAS).
- 2. Find an unknown measure of an angle of a triangle when we know all its three side lengths (Side-Side-Side SSS).
- In order to do that, rewrite the Law of Cosines to show the cosine of the angle, then apply the function arccos to both sides of the equality.
- Since cos and arccos are inverse functions, arccos(cos(∡B)) = ∡B:
- cos(∠B) = (a2 + c2 - b2) / (2 × a × c)
- arccos(cos(∠B)) = arccos((a2 + c2 - b2) / (2 × a × c))
- m(∡B) = arccos((a2 + c2 - b2) / (2 × a × c))
The same process could be used to demonstrate the formulas for the measures of the other two angles, ∠A and ∠C.
Calculations examples
- Example 1: Calculating a side length when the opposite angle and the other two sides are known (SAS)
- In a triangle where two sides' lengths are given, a = 5, c = 7, and the measure of the angle between them is ∡B = 54°, calculate the length of side b:
- Feed the given data into one of the three equations of the Law of Cosines.
- Use a calculator to calculate the function cos(54°).
- b2 = 52 + 72 - 2 × 5 × 7 × cos(54°) = 74 - 70 × 0.5877852522925 = 74 - 41.144967660475 = 32.855032339525
- The square root of b2 = 32.855032339525 is b = 5.7319309433667 ≈ 5.73.
- Example 2: Calculating the measure of an angle when all three side lengths are known (SSS)
- In a triangle where the sides' lengths are a = 5, b = 5.73 and c = 7, calculate the measure of the angle ∠B:
- Feed the given data into one of the three equations rearranged for angles of the Law of Cosine.
- Remember: a2 - b2 = (a + b) × (a - b).
- Use a calculator to calculate the function arccos().
- m(∡B) = arccos((52 + 72 - 5.732) / (2 × 5 × 7)) = arccos((25 + (7 + 5.73) × (7 - 5.73) / 70) = arccos((25 + 12.73 × 1.27) / 70) = arccos(41,1671 / 70) = arccos(0.5881014285714) = 53.9776047480019° ≈ 53.98°.