Law of Cosines

The Law of Cosines

Label every triangle the standard way: each side gets the lowercase letter of the vertex it is opposite. So side $a$ is opposite angle $A$, side $b$ is opposite angle $B$, and side $c$ is opposite angle $C$.

Compare it with the Pythagorean Theorem. If $C = 90^\circ$, then $\cos C = 0$ and the extra term disappears: $c^2 = a^2 + b^2$. The Law of Cosines is the Pythagorean Theorem plus a correction for angles that are not right angles.

Here is when to use it:

Try setting up the formula yourself.

Finding a Side: Two Sides and the Included Angle

If you know two sides and the angle between them (SAS), substitute into the formula. The answer is $c^2$, so finish with a square root.

That answers the question from the start. The distance 249 ft is shorter than $\sqrt{310^2 + 240^2} \approx 392$ ft, as it should be for an acute angle.

For exact answers you need the cosines of the special angles. Remember that for supplementary angles, $\cos(180^\circ - \theta) = -\cos\theta$.

Finding an Angle: Three Sides

If you know all three sides (SSS), solve the Law of Cosines for the cosine of the angle you want, then use $\cos^{-1}$.

Only the square of the side opposite the angle is subtracted. If the result is positive, the angle is acute. If it is negative, the angle is obtuse.

This also covers "viewing angle" problems. If you stand 20 ft from one end of a banner and 24 ft from the other end, and the banner is 10 ft wide, the angle you see it at is the angle opposite the 10 ft side.

Distances, Bearings and Forces

Whenever two distances and the angle between them are measurable but the third distance is not, you have an SAS triangle: surveying, navigation, satellites, or combining forces.

Bearings

A bearing is an angle measured clockwise from north, written with three digits. A bearing of $040^\circ$ means $40^\circ$ to the east of north. The angle between two paths that leave the same point is the difference of their bearings.

Combining forces

Two forces $F_1$ and $F_2$ pull on the same point and make an angle $\theta$ between them. The resultant $R$ is the diagonal of the parallelogram they form. The triangle $O$-$P$-$R$ uses $F_1$ and a copy of $F_2$ placed head to tail, so the angle inside the triangle is $180^\circ - \theta$, not $\theta$.

Parallelograms, Diagonals and Chords

Many geometry figures hide a triangle where you know two sides and the included angle. The trick is to find that triangle.

Diagonals of a parallelogram

Consecutive angles of a parallelogram are supplementary, so a parallelogram with a $70^\circ$ angle also has $110^\circ$ angles. Each diagonal is the side opposite one of these angles in a triangle made of two adjacent sides.

Diagonals that meet at a known angle

The diagonals of a parallelogram bisect each other. If you know the diagonals and the angle where they cross, work with the half-diagonals. (The exercise on this topic calls the shape a rhomboid: a parallelogram whose adjacent sides are unequal and whose angles are not right angles.) The two crossing angles are $\theta$ and $180^\circ - \theta$, so two different side lengths appear.

Chords of a circle

Two radii and a chord form an isosceles triangle, with the central angle between the two equal sides. For radius $r$ and central angle $\theta$:

Equations and Proofs with the Law of Cosines

Sometimes the unknown is hidden in an expression like $x$, $x+1$, $x+2$, or you have to show that a relationship holds for every triangle. The Law of Cosines becomes an equation or an algebra tool.

Matching an algebraic relation to an angle

If a problem gives a relation like $a^2 = b^2 + c^2 + k\,bc$, compare it term by term with $a^2 = b^2 + c^2 - 2bc\cos A$. The coefficients of $bc$ must match, so $-2\cos A = k$ and $\cos A = -\tfrac{k}{2}$. For example, $a^2 = b^2 + c^2 + bc$ gives $\cos A = -\tfrac12$, so $A = 120^\circ$. Use the exact-value table to read the angle.

Known perimeter and one angle

When you know one side $c$, the opposite angle $C$, and the perimeter, you know $a + b$. Rewrite the squares using $a^2 + b^2 = (a+b)^2 - 2ab$.

Proving relationships

To prove something that holds in every triangle, write the Law of Cosines for two or three sides and combine the equations.

Two Triangles, One Shared Side

For the hardest problems, one triangle isn't enough. Find a segment that belongs to two triangles. The Law of Cosines in each triangle gives two expressions for the same quantity, or an angle from one triangle you can use in the other.

A median

The median goes from a vertex to the midpoint of the opposite side. First find an angle of the whole triangle, then reuse it in the smaller triangle that contains the median.

A cyclic quadrilateral

If a quadrilateral is inscribed in a circle, its opposite angles are supplementary: $B + D = 180^\circ$. So $\cos D = -\cos B$. Use the diagonal $AC$ as the shared side of triangles $ABC$ and $ADC$.

Two intersecting circles

The line through the centers is the perpendicular bisector of the common chord. In the triangle formed by the two centers and one intersection point $P$, the Law of Cosines gives an angle. Then the height from $P$ to the line of centers is half the chord.

Diagonals of a cube

For a cube with edge $a$, the diagonal of a face is $a\sqrt2$ (Pythagorean Theorem), and the space diagonal is $\sqrt{a^2 + (a\sqrt2)^2} = a\sqrt3$. Two space diagonals meet at the center of the cube, each cut into halves of length $\tfrac{a\sqrt3}{2}$. Those two halves and an edge of the cube form a triangle you can solve with the Law of Cosines.

Summary