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The 30-60-90 Triangle

How to use
  1. ▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original figure. Learn more about the controls
  2. Drag AA or CC to turn and resize the triangle; the readout under the canvas compares BCBC with AB/2AB/2 and ACAC with AB3/2AB\sqrt3/2. Learn more about why it holds
  3. Labels toggles the point names, Measures writes each marked angle's size beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene
Proof?Reflect B in the leg AC to B′. The triangles ACB and ACB′ are congruent (SAS), so the angle at A doubles to 60° and the angle at B′ is 60° like the angle at B: ABB′ has three 60° angles, so it is equilateral. Then BB′ = AB and BC = BB′/2 = AB/2, and Pythagoras gives AC = AB·√3/2. Drag A or C.30-60-90 triangle
Display
Drag A or C to turn and resize the triangle
αβCAB
BC = 2.17, AB/2 = 2.17, AC = 3.75, AB·√3/2 = 3.75, ∠A = 30°
Step 1 / 6

Step explanations

1A right triangle with angles 30° and 60°
Given
ABC has a right angle at C, the angle α = 30° at A and the angle β = 60° at B. To prove: BC = AB/2, the side opposite 30° is half the hypotenuse, and AC = AB · √3/2.
Drag A or C: the angle at A stays 30°. Learn more about the statement

Getting Started with the Visualizer
The Statement
Reflecting the Triangle
Side, Angle, Side
An Equilateral Triangle
Half the Hypotenuse
Why It Holds
Dragging the Figure
Consequences



Getting Started with the Visualizer

DemoA run through the proof player
Step 0 of 7
Step through the proof, then drag AA or CC to turn and resize the triangle, and watch the readout: BCBC stays equal to AB/2AB/2 and ACAC to AB3/2AB\sqrt3/2, while the angle at AA stays 30°30°. The picture below is the player with its parts numbered; the other tabs list what can be dragged, what each control does and what the step log shows.
Opening scene: the controls
1234567
1Proof — the theorem loaded in the player; the ? gives its idea in one line
2Display — Labels shows the point names, Measures writes each marked angle's size beside its arc
3Draggable points — A and C move freely and turn or resize the triangle; the hollow points B and B′ follow
4Readout — the line under the figure: BC beside AB/2, AC beside AB·√3/2, and the angle at A, live while you drag
5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original figure
6Speed and step counter — how fast Play advances, and which of the 6 steps is on screen
7Step explanations — every step reached so far, with its justification tag and a link to the section that explains it; click an entry to jump back to it
Proof complete: step 6 of 6
12345
1The mirror image — ACB′, the triangle reflected in its long leg AC
2The equilateral triangle — ABB′, with two ticks on AB and AB′
3The short leg — BC, half of BB′, one tick like CB′
4Justification tags — Given, Construction, SAS, Equilateral triangle, Halving, Pythagorean theorem
5Replay — at the last step Play becomes Replay and Next is off

The Statement

A 30-60-90 triangle is a right triangle with acute angles 30°30° and 60°60°. In the player the right angle is at CC, the 30°30° angle at AA and the 60°60° angle at BB. The theorem gives its sides:

BC=AB2,AC=32 AB,BC:AC:AB=1:3:2BC = \frac{AB}{2}, \qquad AC = \frac{\sqrt3}{2}\,AB, \qquad BC : AC : AB = 1 : \sqrt3 : 2


The side opposite the 30°30° angle is half the hypotenuse. This is one of the two special right triangles, together with the 45-45-90 triangle on the square diagonals page, and it is the source of the exact values sin⁡30°=12\sin 30° = \tfrac12 and cos⁡30°=32\cos 30° = \tfrac{\sqrt3}{2}.

The trick is that a 30-60-90 triangle is half of an equilateral triangle. In the player AA and CC are free and BB is placed on the perpendicular at CC so that the angle at AA stays 30°30°; the readout compares the sides with AB/2AB/2 and AB3/2AB\sqrt3/2.
αβCAB
Opening scene, frozen

The right triangle ABC with α = 30° at A and β = 60° at B.


Reflecting the Triangle

Extend the leg BCBC beyond CC to a point B′B' with CB′=CBCB' = CB, and join AA to B′B'. Because the angle at CC is right, B′B' is the mirror image of BB in the line ACAC, and BB, CC, B′B' lie on one line. The triangle and its reflection together make the big triangle ABB′ABB'.
αβCABB′
Reflecting the triangle, frozen

B reflected in AC to B′: CB = CB′.


Side, Angle, Side

The triangles ACBACB and ACB′ACB' share the side ACAC, have CB=CB′CB = CB', and have right angles at CC between those sides. By SAS they are congruent, so

AB′=AB,∠CAB′=30°,∠AB′C=60°AB' = AB, \qquad \angle CAB' = 30°, \qquad \angle AB'C = 60°
αβαβCABB′
Side, angle, side, frozen

ACB and ACB′ congruent by SAS: 30° and 60° copied.


An Equilateral Triangle

Now look at the big triangle ABB′ABB'. Its angle at AA is made of two 30°30° angles, so it is 60°60°; its angles at BB and B′B' are 60°60° each. A triangle with three equal angles is equilateral:

AB=BB′=AB′AB = BB' = AB'
αβαβCABB′
An equilateral triangle, frozen

Three 60° angles: ABB′ is equilateral.


Half the Hypotenuse

CC is the midpoint of BB′BB', since CB=CB′CB = CB'. So

BC=BB′2=AB2BC = \frac{BB'}{2} = \frac{AB}{2}


The short leg, opposite the 30°30° angle, is half the hypotenuse.
αβαβCABB′
Half the hypotenuse, frozen

BC is half of BB′, and BB′ = AB.


Why It Holds

The long leg comes from the Pythagorean theorem in the right triangle ABCABC:

AC2=AB2−BC2=AB2−AB24=34 AB2,AC=32 ABAC^2 = AB^2 - BC^2 = AB^2 - \frac{AB^2}{4} = \frac34\,AB^2, \qquad AC = \frac{\sqrt3}{2}\,AB


So the sides are in the ratio 1:3:21 : \sqrt3 : 2. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, a construction, SAS, the equilateral triangle, halving, and the Pythagorean theorem.

The converse also holds: if one leg of a right triangle is half the hypotenuse, the angle opposite it is 30°30°. The same reflection gives a triangle with three equal sides, hence three 60°60° angles, and the angle at AA is half of one.
αβαβCABB′
The conclusion, frozen

Pythagoras: AC = AB · √3/2, the ratio 1 : √3 : 2.


Dragging the Figure

Drag AA or CC: the triangle turns and grows or shrinks, BB and its mirror image B′B' follow, and the readout shows BC=AB/2BC = AB/2 and AC=AB3/2AC = AB\sqrt3/2 at every position, with the angle at AA fixed at 30°30°. Only the scale changes, never the shape.

The player keeps the long leg long enough for the marks and both the triangle and its mirror image inside the frame; a drag that would push a point out is refused.
αβαβCABB′
After dragging A, frozen

A moved in and up: the triangle turned and shrank, the ratio 1 : √3 : 2 unchanged.


Consequences

• Exact trigonometric values. With hypotenuse 1 the legs are 12\tfrac12 and 32\tfrac{\sqrt3}{2}: sin⁡30°=cos⁡60°=12\sin 30° = \cos 60° = \tfrac12, cos⁡30°=sin⁡60°=32\cos 30° = \sin 60° = \tfrac{\sqrt3}{2}, tan⁡30°=13\tan 30° = \tfrac{1}{\sqrt3}, tan⁡60°=3\tan 60° = \sqrt3.
• The height of an equilateral triangle. An equilateral triangle of side ss has height 32s\tfrac{\sqrt3}{2}s and area 34s2\tfrac{\sqrt3}{4}s^2.
• The midpoint of the hypotenuse. The median from the right angle is half the hypotenuse (midpoint of the hypotenuse); in a 30-60-90 triangle it is also equal to the short leg, cutting off an equilateral triangle at the 60°60° vertex.
• The regular hexagon. Half of each of the six equilateral triangles of a regular hexagon is a 30-60-90 triangle, so the distance from the centre to a side is 32\tfrac{\sqrt3}{2} times the side.
• Special right triangles. Together with the 45-45-90 triangle of the square diagonals page, it is one of the two right triangles whose sides and angles are both known exactly without tables.