Geometry: An Interactive Journey to Mastery

Example 1
Find sin 120° and cos 120°.
Solution
'UDZDVNHWFKͼ௘6HHFigure 16.2௘ͽ
We see half of an equilateral triangle.
Thus, cos 120° ^12 ͼ௘WKHGLVSODFHPHQWLVLQWKHQHJDWLYHGLUHFWLRQ௘ͽ
and an application of the Pythagorean theorem gives sin 120° 23.

Example 2
Find sin 135° and cos 135°.
Solution
'UDZDVNHWFKͼ௘6HHFigure 16.3௘ͽ
We have a right isosceles triangle. An application of the Pythagorean
theorem gives each side length of that triangle to be^12.
Thus, sin  135°^12 and cos  135°^12.
Example 3
Find all values for x satisfying sin x 12.
Solution
A sketch shows that there are essentially two locations for which
sin x 12 .ͼ௘6HHFigure 16.4௘ͽ
We see half equilateral triangles, so these two locations are based
on an angle of 30°.
Thus, x = 30° or 150° or the addition of multiples of 360° to these
two angles.

1

60°

120°

Figure 16.2

1

45°

Figure 16.3

1 11

2 12

Figure 16.4