Law of Cosines

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Law of Cosines: meaning, method and solved example (Geometry) – Maths PDF


In short: The law of cosines stretches the Pythagorean theorem to every triangle, producing a missing edge from two edges and the corner between them, or a corner from three edges.

GeometryTrigonometryG-SRT.D.10

What is Law of Cosines?

The law of cosines is the Pythagorean theorem plus a correction term. For edges \(a\), \(b\), \(c\) and the corner \(C\) facing edge \(c\): \(c^2 = a^2 + b^2 – 2ab\cos C\).

At \(C = 90^{\circ}\) the cosine is 0, the correction disappears, and \(c^2 = a^2 + b^2\) returns. A narrow corner subtracts something, so \(c\) shrinks; a wide corner has a negative cosine, so \(c\) grows. Rearranged, the formula yields a corner: \(\cos C = \frac{a^2 + b^2 – c^2}{2ab}\).

How to spot it

  • Two edges of a triangle and the corner squeezed between them are given, and the third edge is requested.
  • Every edge length is known and a corner measure is wanted, a job for the law of cosines.
  • No edge is paired with the corner across from it, so the sines version has nothing to start from.
  • A navigation story gives two distances from a starting point and the turn between the two paths.

The method

  1. Label the unknown side \(c\) and the angle facing it \(C\); the other two sides are \(a\) and \(b\).
  2. Substitute into \(c^2 = a^2 + b^2 – 2ab\cos C\), computing the cosine with the calculator in degree mode.
  3. Simplify to a number for \(c^2\), then take the square root.
  4. To find an angle instead, use \(\cos C = \frac{a^2 + b^2 – c^2}{2ab}\) and the inverse cosine key.
CAB55°b = 14a = 9c ≈ 11.5c² = 9² + 14² – 2(9)(14)cos 55°c² ≈ 132.46
Sides 9 and 14 around a 55° angle force the third side to about 11.5.

Law of Cosines: solved example

Problem

Two hiking trails leave a trailhead at a \(55^{\circ}\) angle to each other. One runs 9 miles to a lake and the other runs 14 miles to a ridge. How far apart are the lake and the ridge? Use \(\cos 55^{\circ} \approx 0.5736\).

Solution

The angle between the 9-mile and 14-mile sides faces the unknown distance \(c\): \(c^2 = 9^2 + 14^2 – 2(9)(14)\cos 55^{\circ} = 81 + 196 – 252(0.5736)\).

That is \(277 – 144.55 = 132.45\), and \(\sqrt{132.45} \approx 11.5\). The lake and ridge are about 11.5 miles apart.

Try it

Question 1

A triangular garden bed has edges of 5 m, 7 m and 8 m. Find the corner across from the 8 m edge.

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\(\cos C = \frac{25 + 49 – 64}{2(5)(7)} = \frac{10}{70} = \frac{1}{7}\), so \(C \approx 81.8^{\circ}\).

Question 2

What happens to the formula when the corner \(C\) is exactly square?

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\(\cos 90^{\circ} = 0\), so \(c^2 = a^2 + b^2\): the Pythagorean theorem returns.

Law of Cosines vs. Pythagorean Theorem

Pythagoras demands one corner of exactly \(90^{\circ}\). The law of cosines accepts any triangle and swallows Pythagoras as the special case with a square corner. If your triangle has such a corner, the shortcut is enough; otherwise you need the full formula with its correction term.

Pythagorean Theorem

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