Exercise 10 Page 916

We want to find all the angles whose tangent is $\text{-}\sqrt{3}.$ Since we know the value of tangent but not the value of the angle, the process of finding the missing value uses an inverse trigonometric function. First, we need to use the definition of tangent to find the appropriate reference angle. The tangent of an angle is defined as the ratio of the sine to the cosine of the angle. Note that we will actually find the opposite of the desired angle because all of the reference angles have positive values for sine and cosine while the given tangent value is negative.

	$\sin \theta$	$\cos \theta$	$\frac{\sin \theta }{\cos \theta }$	$\tan \theta$
$\theta =30^{\circ }$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$	$\frac{\sqrt{3}}{3}$
$\theta =45^{\circ }$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}$	$1$
$\theta =60^{\circ }$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}$	$\sqrt{3}$
$\theta =90^{\circ }$	$1$	$0$	$\frac{1}{0}$	Undefined

The tangent of an angle that measures $60^{\circ }$ is $\sqrt{3}.$ Let's now construct a $30^{\circ }\text{-}60^{\circ }\text{-}90^{\circ }$ 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 $\text{shorter}$ $\text{leg}$ is half the length of the $\text{hypotenuse},$ and the length of the $\text{longer}$ $\text{leg}$ is $\sqrt{3}$ times the length of the $\text{shorter}$ $\text{leg}.$

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 $30^{\circ }\text{-}60^{\circ }\text{-}90^{\circ }$ triangle on a unit circle with the measurements found above in these quadrants. We will place the $60^{\circ }$ angle at the origin.

Finally, we will use these triangles to locate two angles whose tangent is $\text{-}\sqrt{3}.$ Keep in mind that a full turn measures $360^{\circ }$ and a half turn measures $180^{\circ }.$ Therefore, we will subtract $60$ from $180$ and from $360.$ Let's show the obtained angles on the coordinate plane. We found that $\tan \left(\text{-}\sqrt{3}\right)$ can be either $120^{\circ }$ or $300^{\circ }.$ Finally, keep in mind that if we add or subtract a multiple of $360^{\circ },$ the terminal side of the angle will be in the same position. Therefore, the tangent of the resulting angles will also be $\text{-}\sqrt{3}.$