Identities and Area in a Triangle

What this move does

The three angles of a triangle satisfy $A + B + C = π$ , so once you know two of them, the third is fixed. Almost every identity in a triangle rests on this one fact:

C = π - (A + B) ⇓ sin C = sin (A + B) cos C = - cos (A + B) tan C = - tan (A + B)

Half-angles work the same way: $\frac{C}{2} = \frac{π}{2} - \frac{A + B}{2}$ , so $sin \frac{C}{2} = cos \frac{A + B}{2}$ .

Use $C = π - (A + B)$ (or its half-angle form) to eliminate one angle.
Expand with the angle sum formulas and the sum-to-product formulas.
Tidy up and, when needed, turn $A + B$ back into $π - C$ so that the result is symmetric in the three angles.

The lab above draws the same idea as a fan: the three pieces $A$ , $B$ and $C$ fit together into exactly a straight angle.

Predict first

In $△ A B C$ , $tan A = 1$ and $tan B = 2$ . What is $tan C$ ?

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

Work through one

Proof: tan A + tan B + tan C = tan A tan B tan C

Step 1 / 5

Look at the numbers first
$A = 7 0^{\circ}$ , $B = 5 0^{\circ}$ , $C = 6 0^{\circ}$ . The two sides come out the same. What we have to prove is that this happens for every triangle.

$tan A + tan B + tan C = ? tan A tan B tan C$
Eliminate one angle
Expand with the angle sum formula
Clear the denominator and rearrange
Right triangles excepted

Area: one number, five ways to write it

$S = \frac{1}{2} ab sin C$ is the starting point. Replace the sides by $2 R sin A$ and so on, or cut the triangle into three pieces from the incenter, and you get the other forms.

Starting from ½ab sin C

Step 1 / 5

Two sides and the angle between them
Take $a$ as the base; the height is $b sin C$ .

$S = \frac{1}{2} a \cdot b sin C$
Replace sin C using the law of sines
Now replace the sides instead
Cut into three from the incenter
Sides only: Heron’s formula

When it fails

Break it

Counterexample lab: remove one condition

Conclusion: $tan A + tan B + tan C = tan A tan B tan C$

The three angles come from a real triangle, and $C$ is determined by $A$ and $B$ . Every identity holds, whichever you pick.

Practice

Exercise 1Basic

In $△ A B C$ , $a = 4$ , $b = 5$ and $C = 3 0^{\circ}$ . Find the area $S$ .

Exercise 2Advanced

A triangle has sides $13$ , $14$ and $15$ . Find its circumradius $R$ .

Exercise 3Advanced

In $△ A B C$ , $tan A = 2$ and $tan B = 3$ . Find angle $C$ in degrees.

Exercise 4Challenge

$△ A B C$ satisfies $sin^{2} A + sin^{2} B + sin^{2} C = 2$ . Find its largest angle (in degrees).

What this move does

Work through one

Proof: tan A + tan B + tan C = tan A tan B tan C

Area: one number, five ways to write it

Starting from ½ab sin C

When it fails

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Concluding A = B straight from sin 2A = sin 2B