Transcript new angles that are 30

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.
a b  c
2
2
1  b  2
2
2
 b2  3
b 3
2
2
• This 30-60-90 right triangle can give us the
trigonometric function values of 30° and 60°. We
first do the 30° angle.
opp 1
sin30 

hyp 2
3
adj

cos30 
2
hyp
1
opp

tan30 
adj
3
• We now do the 60° angle.
opp
3
sin60 

2
hyp
adj 1

cos60 
hyp 2
3
opp

 3
tan60 
1
adj
• This leads us to some important values on the unit
circle.
• Recall that on the unit circle we have …
(a, b)  (cos x,sin 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 …

3
cos 
6
2

1
sin 
6 2
• Moving around the unit circle with reference angles
of π/6 we have …
Example 1:
Find cos 5π/6
• Since cos x is equal to the first coordinate of the
point we have …
5
3
cos

6
2
Example 2:
Find sin 7π/6
• Since sin x is equal to the second coordinate of the
point we have …
7
1
sin

6
2
Example 3:
Find tan (-7π/6)
• Since tan x is equal to b/a we have …
 7
tan  
 6



1
2
3

2
1
2
1
 

2
3
3
• In similar fashion, moving around the unit circle with
reference angles of π/3 (60°) we have …
Example 1:
Find cos 4π/3
• Since cos x is equal to the first coordinate of the
point we have …
4
1
cos

3
2
Example 2:
Find sin -2π/3
• Since sin x is equal to the second coordinate of the
point we have …
 2
sin  
 3
3


2

Example 3:
Find tan (2π/3)
• Since tan x is equal to b/a we have …
3
 2 
2
tan 


 3  1
2
3 2

   3
2
1