The Laws of Sines and Cosines

Try this

Drag $A$ while keeping it above chord $B C$ : neither $∠ A$ nor side $a$ changes, because inscribed angles on the same arc are equal.
Drag $B$ or $C$ to change side $a$ : now $∠ A$ changes too, yet “ $a \div sin A$ ” always equals the diameter $2 R$ .
Drag $A$ below chord $B C$ . Now $∠ A$ is obtuse, and $∠ B D C$ becomes its supplement (the dashed arc). Has the ratio changed?
Click “side $b$ ” and then “side $c$ ”: each side gets its own diameter, and the answer is the same $2 R$ every time.

Put these four together and you have the law of sines:

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

where $R$ is the circumradius. It is far more precise than “the larger side faces the larger angle”: the ratio of the sides is exactly the ratio of the sines of the opposite angles, $a : b : c = sin A : sin B : sin C$ .

Predict first

The circumradius of $△ A B C$ is $R = 4$ and $∠ A = 15 0^{\circ}$ . How long is side $B C$ ?

Choose an answer first; then you can see the result.

Why: draw a diameter in the circumcircle

Proof: a = 2R sin A

Step 1 / 6

What we want to prove
The triangle is inscribed in a circle of radius $R$ . Right now all three ratios show the same number; we want a reason that works for every triangle.

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = ? 2 R$
Draw the diameter through B
An angle in a semicircle is a right angle
Replace ∠D by ∠A
When A is obtuse
All three sides alike

The law of cosines: Pythagoras, corrected

The law of sines links a side with its opposite angle. When you know two sides and the angle between them, you need the law of cosines for the third side: drop the altitude from $A$ to line $B C$ , which cuts a right triangle out of the oblique one.

Proof: c² = a² + b² − 2ab cos C

Step 1 / 5

Draw the altitude AH
In $△ A C H$ , $A H$ is the side opposite $∠ C$ and $C H$ is the adjacent side.

$A H = b sin C, C H = b cos C$
The right triangle AHB
Expand and use sin² + cos² = 1
Obtuse angles work too
A right angle: the correction vanishes

Used the other way round: when all three sides are known, $cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}$ . The sign of the numerator alone tells you whether $C$ is acute, right or obtuse.

Remove the right angle

Break it

Counterexample lab: remove one condition

Conclusion: $c^{2} = a^{2} + b^{2}$

$C$ is locked at a right angle. Change $a$ and $b$ as much as you like: the correction stays $0$ .

Practice

Exercise 1Basic

In $△ A B C$ , $a = 9$ and $A = 15 0^{\circ}$ . Find the circumradius $R$ .

Exercise 2Basic

A triangle has sides $4$ , $5$ and $6$ . Find the cosine of its largest angle. Is the triangle acute, right or obtuse?

Exercise 3Advanced

In $△ A B C$ , $sin A : sin B : sin C = 5 : 7 : 8$ . Find $cos B$ .

Exercise 4Challenge

In $△ A B C$ , $b = 23$ , $c = 2$ and $C = 3 0^{\circ}$ . Find all possible values of side $a$ (separate them with commas).

Try this

Why: draw a diameter in the circumcircle

Proof: a = 2R sin A

The law of cosines: Pythagoras, corrected

Proof: c² = a² + b² − 2ab cos C

Remove the right angle

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Finding an angle from its sine and keeping only the acute one