Grade 10

Trigonometric relations in triangles

The law of sines and law of cosines, connecting a triangle's side lengths to its angles.

IntuitionVisual intuition

A triangle has six measurements — three side lengths and three angles — but they are not independent: fixing some of them pins down all the rest. If you know two sides and the angle between them, the law of cosines hands you the third side directly, generalizing the Pythagorean theorem to triangles that are not right-angled. If instead you know one side and its opposite angle, the law of sines lets you scale that relationship to every other side-angle pair, and even tells you the radius of the circle that passes through all three vertices.

A triangle inscribed in its circumcircle of radius

R = a

: by the Law of Sines, each side length equals

2R\sin

of its opposite angle.

SchoolPrecise statements

Definition: Notation for a triangle's sides, angles, and circles

In triangle $ABC$ , side $a = BC$ is opposite vertex $A$ , side $b = CA$ is opposite vertex $B$ , and side $c = AB$ is opposite vertex $C$ . Let $R$ be the radius of the circumscribed circle (through all three vertices), $r$ the radius of the inscribed circle (tangent to all three sides), and $p = \frac{a+b+c}{2}$ the semiperimeter.

c^2 = a^2 + b^2 - 2ab\cos C

This is the law of cosines: $c^2 = a^2 + b^2 - 2ab\cos C$ . Here $C$ is the angle at vertex $C$ , opposite side $c$ . When $C = 90°$ , $\cos C = 0$ and the formula reduces exactly to the Pythagorean theorem $c^2 = a^2+b^2$ — the law of cosines is a generalization that works for any angle, not just right angles.

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R

This is the law of sines: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$ . It says the ratio of a side to the sine of its opposite angle is the same for all three sides, and that common value is exactly the diameter $2R$ of the circumscribed circle — a single number that captures the overall scale of the triangle.

Four ways to compute a triangle's area
Formula	Needs
$S = \frac{1}{2}ab\sin C$	Two sides and the included angle
$S = \frac{abc}{4R}$	All three sides and the circumradius $R$
$S = pr$	The semiperimeter $p$ and the inradius $r$
$S = \sqrt{p(p-a)(p-b)(p-c)}$	All three sides only (Heron's formula)

UndergraduateTheorems and proofs

Theorem: The law of cosines

In any triangle $ABC$ with sides $a,b,c$ opposite $A,B,C$ , $c^2 = a^2 + b^2 - 2ab\cos C$ .

Why is it true?

Placing the triangle in coordinates with the known angle at the origin turns the unknown side into a distance formula, and expanding that distance formula automatically produces the extra cosine term beyond the Pythagorean sum.

Proof

Place vertex $C$ at the origin, vertex $A$ at $(b, 0)$ along the positive $x$ -axis (since $CA = b$ ), and vertex $B$ at $(a\cos C, a\sin C)$ — this is the point at distance $a$ from $C$ (since $CB = a$ ) making angle $C$ with the segment $CA$ .

The side $c = AB$ is the distance between $A=(b,0)$ and $B=(a\cos C, a\sin C)$ , so $c^2 = (b - a\cos C)^2 + (a\sin C)^2$ .

Expanding the square: $c^2 = b^2 - 2ab\cos C + a^2\cos^2 C + a^2\sin^2 C$ .

Since $\cos^2 C + \sin^2 C = 1$ , the last two terms combine to $a^2$ , leaving $c^2 = a^2 + b^2 - 2ab\cos C$ , which is exactly $c^2 = a^2 + b^2 - 2ab\cos C$ . Repeating the same argument with a different vertex at the origin gives the analogous formulas for $a^2$ and $b^2$ .

Theorem: The law of sines

In any triangle $ABC$ inscribed in a circle of radius $R$ , $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$ .

Why is it true?

Every inscribed angle subtending the same arc is equal, so replacing an angle by an equal angle at a cleverly chosen point turns the triangle into a right triangle inside the circle, where sine is just a ratio of a chord to the diameter.

Proof

Let $O$ be the circumcenter and draw the diameter through $B$ , meeting the circle again at a point $B'$ . Since $BB'$ is a diameter, the inscribed angle $\angle BCB'$ subtends a semicircle, so $\angle BCB' = 90°$ .

The inscribed angles $\angle BAC$ and $\angle BB'C$ both subtend the same arc $BC$ (or are supplementary if $A$ and $B'$ lie on opposite arcs, in which case their sines are still equal since $\sin(180°-x)=\sin x$ ), so $\angle BB'C = A$ or its supplement, and in either case $\sin(\angle BB'C) = \sin A$ .

In right triangle $BCB'$ (right-angled at $C$ ), the side $BC = a$ is opposite $\angle BB'C$ , and the hypotenuse $BB' = 2R$ is the diameter, so $\sin(\angle BB'C) = \dfrac{a}{2R}$ .

Combining the last two steps, $\sin A = \dfrac{a}{2R}$ , i.e. $\dfrac{a}{\sin A} = 2R$ . Repeating the same construction with the diameter through $A$ or through $C$ gives $\dfrac{b}{\sin B} = 2R$ and $\dfrac{c}{\sin C} = 2R$ , so all three ratios equal $2R$ , proving $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$ .

UndergraduateReal-World Applications and Worked Examples

These two laws are the backbone of triangulation, the technique of finding an unknown distance or position from known angles and a known side. Navigators and surveyors use the law of cosines to compute the distance between two ships or landmarks whose bearings and distances from a common point are known, and GPS and cell-tower positioning rely on the same triangle geometry to locate a receiver from signals with known angles or timing. Civil engineers use the law of sines to find unmeasurable lengths in trusses and roof frames, and land surveyors use Heron's formula to compute the area of an irregular plot from only its three (or more, by triangulating) side lengths, without needing to measure any angle directly.

Example: Distance between two ships

Two ships leave the same port. Ship 1 sails $30$ km along one heading, and ship 2 sails $40$ km along a heading that makes a $60°$ angle with ship 1's heading. Find the distance between the two ships.

Solution

The port and the two ships form a triangle where the two known sides ( $30$ km and $40$ km) meet at the port with a known included angle ( $60°$ ) — exactly the setup for the law of cosines, $c^2 = a^2 + b^2 - 2ab\cos C$ , with $a=30$ , $b=40$ , $C=60°$ , and $c$ the unknown distance between the ships.

Substituting: $c^2 = 30^2 + 40^2 - 2\cdot30\cdot40\cdot\cos 60°= 900+1600-2400\cdot\frac{1}{2}$ .

Computing: $c^2 = 2500 - 1200 = 1300$ , so $c = \sqrt{1300}\approx 36.1$ km.

Example: Area of a triangular plot of land

A triangular plot of land has sides $13$ m, $14$ m, and $15$ m. Find its area.

Solution

Only the three side lengths are known and no angle is measured directly, so Heron's formula $S = \sqrt{p(p-a)(p-b)(p-c)}$ is the right tool, with semiperimeter $p = \frac{13+14+15}{2} = 21$ .

Substituting: $S = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21\cdot 8\cdot 7\cdot 6}$ .

Computing the product inside the root: $21\cdot 8 = 168$ , $7\cdot 6=42$ , and $168\cdot 42 = 7056$ .

So $S = \sqrt{7056} = 84$ . The plot has area $84$ m².

A triangle has sides $a = 8$ , $b = 5$ , and included angle $C = 60°$ . Find the length of side $c$ .

A triangle has sides $a=5$ , $b=12$ , $c=13$ . Which of the following is true about angle $C$ (opposite side $c$ )?

A triangle has side $a = 10$ opposite an angle $A = 30°$ . What is the diameter $2R$ of its circumscribed circle?

A surveyor measures the three sides of a triangular field as $9$ m, $10$ m, and $17$ m, without measuring any angle. Which formula should be used to find the field's area?