30-60-90 Right Triangles

• Consider the following equilateral triangle, with each side having a value of 2.

• Drop a perpendicular segment from the top vertex to the base side.

• The base side has been bisected into two segments of length 1.

• Since the original triangle was equilateral, the base angles are 60° each.

• Since the angle at the top of the triangle has been bisected, the original 60° angle has been split into new angles that are 30° each.

• Consider the new triangle formed on the left side.

• This is a 30-60-90 right triangle.

• We have just proven that in a 30-60-90 triangle the short leg (always opposite the 30° angle) is always half the length of the hypotenuse.

• Name the long leg b and determine its value.

2 2 2 a b c

2 2 21 2 b

2 3 b

3 b

• This 30-60-90 right triangle can give us the trigonometric function values of 30° and 60°. We first do the 30° angle.

oppsin30

hyp

adjcos30

hyp

opptan30

adj

1

2

3

2

1

3

• We now do the 60° angle.

oppsin60

hyp

adjcos60

hyp

opptan60

adj

3

2

1

2

33

1

• This leads us to some important values on the unit circle.

• Recall that on the unit circle we have …

( , )a b (cos ,sin )x x

• Consider the point (a,b) on the 30° ray of a unit circle.

• Since (a,b) = (cos 30°, sin 30°) , we have

• In radian form it would be …

3cos

6 2

1sin

6 2

• Moving around the unit circle with reference angles of π/6 we have …

Example 1:Find cos 5π/6

5 3cos

6 2

• Since cos x is equal to the first coordinate of the point we have …

Example 2:Find sin 7π/6

7 1sin

6 2

• Since sin x is equal to the second coordinate of the point we have …

Example 3:Find tan (-7π/6)

17 2tan6 3

2

• Since tan x is equal to b/a we have …

1 2

2 3

1

3

• In similar fashion, moving around the unit circle with reference angles of π/3 (60°) we have …

Example 1:Find cos 4π/3

4 1cos

3 2

• Since cos x is equal to the first coordinate of the point we have …

Example 2:Find sin -2π/3

2 3sin

3 2

• Since sin x is equal to the second coordinate of the point we have …

Example 3:Find tan (2π/3)

32 2tan

132

• Since tan x is equal to b/a we have …

3 2

2 1 3