The equalateral triangle above has sides each 2 units long and all angles at 60o. It has been halved into 2 right - angled triangles of base 1 unit long. Using Pythagorus:
(ab)2 = (ac)2 + (bc)2
=> (2)2 = (ac)2 + (1)2
=> (ac)2 = 4 - 1 = 3
and (ac) = root 3.
Using the definitions of trigonometrical ratios, summarised in SOHCAHTOA:
sin 30o = bc / ab = 1 / 2
sin 60o = ac / ab = root 3 / 2
cos 30o = ac / ab = root 3 / 2
cos 60o = bc / ab = 1 /2
tan 30o = bc / ac = 1 / root 3
tan 60 = ac / bc = root 3 / 1 = root 3
The isosceles above has equal sides of 1 unit each and the 2 complementary angles each of 45o. Using Pythagoras' Theorem to find side (ac):
(ac)2 = (ab)2 + (bc)2| Trigonometrical ratio: | Surd form: | Approximation |
| sin 30o | 1 / 2 | 0.5 |
| sin 45o | 1 / root 2 | 0.7071 |
| sin 60o | root 3 / 2 | 0.866 |
| cos 30o | root 3 / 2 | 0.866 |
| cos 45 | 1 / root 2 | 0.7071 |
| cos 60o | 1 / 2 | 0.5 |
| tan 30o | 1 / root 3 | 0.5774 |
| tan 45o | 1 | |
| tan 60o | root 3 | 1.7321 |
For any pair of complementary angles these 2 rules apply:
sin xo = cos (90 - x)
and cos xo = sin (90 - x)
| sin 30o = cos 60o |
| sin 60o = cos 30o |
| sin 45o = cos 45o |
| sin 20o = cos 70o |
| cos 41o = sin 49o |
| cos 72o = sin 18o |
| sin 36o = cos 54o |
When the angle is 30o on a right - angled triangle, then the side opposite the 30o angle is half the hypotenuse. This is also true for similar triangles. Sides H, A and O can be any value and O = 1 / 2 H. Therefore, sin 30o = 0.5.