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120^(∘) + n* 360^(∘) and 300^(∘) + n* 360^(∘), for any integer n
sin θ | cos θ | sin θ/cos θ | tan θ | |
---|---|---|---|---|
θ=30^(∘) | 1/2 | sqrt(3)/2 | 12/sqrt(3)2 | sqrt(3)/3 |
θ=45^(∘) | sqrt(2)/2 | sqrt(2)/2 | sqrt(2)2/sqrt(2)2 | 1 |
θ=60^(∘) | sqrt(3)/2 | 1/2 | sqrt(3)2/12 | sqrt(3) |
θ=90^(∘) | 1 | 0 | 1/0 | Undefined |
The tangent of an angle that measures 60^(∘) is sqrt(3). Let's now construct a 30^(∘)-60^(∘)-90^(∘) triangle. Since we will also need to use the unit circle, our right triangle will have a hypotenuse that measures 1.
In this type of special triangle, the length of the shorter leg is half the length of the hypotenuse, and the length of the longer leg is sqrt(3) times the length of the shorter leg.
Next, let's locate this triangle in the coordinate plane. Since the given value is negative, our solutions will be located in the quadrants that have negative tangent values — Quadrant II and Quadrant IV.
We can draw a 30^(∘)-60^(∘)-90^(∘) triangle on a unit circle with the measurements found above in these quadrants. We will place the 60^(∘) angle at the origin.
Finally, we will use these triangles to locate two angles whose tangent is - sqrt(3). Keep in mind that a full turn measures 360^(∘) and a half turn measures 180^(∘). Therefore, we will subtract 60 from 180 and from 360. 180- 60&= 120^(∘) 360- 60&= 300^(∘) Let's show the obtained angles on the coordinate plane.
We found that tan(- sqrt(3)) can be either 120^(∘) or 300^(∘). Finally, keep in mind that if we add or subtract a multiple of 360^(∘), the terminal side of the angle will be in the same position. Therefore, the tangent of the resulting angles will also be - sqrt(3). 120^(∘)+ n * 360^(∘) and 300^(∘)+ n * 360^(∘), wheren is any integer