Question

What is the value of $\tan 120,135$ and $150\;$ degrees?

Answer 1

To solve these questions where one is required to find the values of a certain trigonometric ratio with the help of the unit circle, one simply has to look up the values for certain angles in the quadrants of the unit circle for the given trigonometric ratio.

Complete step by step solution:
It is given in the question that,
We have to find the unit circle values of $\tan 120,135$ and $150\;$ degrees.
A unit circle can be defined as a circle that is divided into quadrants and contains angles and their signs which makes it easier to associate a sign with a specific trigonometric ratio. The circle is divided into quadrants that cover all the angles.
The first quadrant covers angles from ${{0}^{\circ }}$ to ${{90}^{\circ }}$ , the second quadrant contains angles from ${{90}^{\circ }}$ to ${{180}^{\circ }}$ , the third quadrant consists of angles from ${{180}^{\circ }}$ to ${{270}^{\circ }}$ , and the last quadrant, that is the fourth quadrant contains angles from ${{270}^{\circ }}$ to ${{360}^{\circ }}$ .
The signs associated with every quadrant can be given by the ASCT rule, which means that all the trigonometric ratios are positive in the first quadrant, sine and cosecant are positive in the second quadrant, cosine and secant are positive in the third quadrant and the ratios tangent and cotangent are positive in the fourth quadrant.
Now, to find the values of $\tan 120,135$ and $150\;$ degrees from the unit circle, first, calculate the values of sine and cosine of the given angles and then use the formula $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ to get the answer.
For the angle ${{120}^{\circ }}$ :
This angle comes in the second quadrant; therefore, the sine value of the angle will be positive and the cosine value will be negative.
We know, that $\sin {{120}^{\circ }}=\dfrac{\sqrt{3}}{2}$ and $\cos {{120}^{\circ }}=\dfrac{-1}{2}$
So, $\tan {{120}^{\circ }}=\dfrac{\sin {{120}^{\circ }}}{\cos {{120}^{\circ }}}$
Now, substituting values, we get,
$\Rightarrow \tan {{120}^{\circ }}=\dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{-1}{2}}$
Simplifying the above expression, we get,
$\Rightarrow \tan {{120}^{\circ }}=-\sqrt{3}$
For the angle ${{135}^{\circ }}$ :
This angle comes in the second quadrant as well, therefore the sine value of the angle will be positive, and the cosine value will be negative.
We know, that $\sin {{135}^{\circ }}=\dfrac{\sqrt{2}}{2}$ and $\cos {{135}^{\circ }}=\dfrac{-\sqrt{2}}{2}$
So, $\tan {{135}^{\circ }}=\dfrac{\sin {{135}^{\circ }}}{\cos {{135}^{\circ }}}$
Now, substituting values, we get,
$\Rightarrow \tan {{135}^{\circ }}=\dfrac{\dfrac{\sqrt{2}}{2}}{\dfrac{-\sqrt{2}}{2}}$
Simplifying the above expression, we get,
$\Rightarrow \tan {{135}^{\circ }}=-1$
For the angle ${{150}^{\circ }}$ :
This angle comes in the second quadrant also, therefore the sine value of the angle will be positive, and the cosine value will be negative.
We know, that $\sin {{150}^{\circ }}=\dfrac{1}{2}$ and $\cos {{150}^{\circ }}=\dfrac{-\sqrt{3}}{2}$
So, $\tan {{150}^{\circ }}=\dfrac{\sin {{150}^{\circ }}}{\cos {{150}^{\circ }}}$
Now, substituting values, we get,
$\Rightarrow \tan {{150}^{\circ }}=\dfrac{\dfrac{1}{2}}{\dfrac{-\sqrt{3}}{2}}$
Simplifying the above expression, we get,
$\Rightarrow \tan {{150}^{\circ }}=-\dfrac{1}{\sqrt{3}}$
This can also be written as $\tan {{150}^{\circ }}=-\dfrac{\sqrt{3}}{3}$ after rationalizing the denominator.
So, the unit circle values of the given angles in the question are as follows:
$\tan {{120}^{\circ }}=-\sqrt{3}$ , $\tan {{135}^{\circ }}=-1$ and $\tan {{150}^{\circ }}=-\dfrac{\sqrt{3}}{3}$

Note: While solving these questions, one must have a basic knowledge of finding the values of the different trigonometric ratios for various angles by using the basic and fundamental properties and laws of trigonometry. Also, one should pay attention to the sign that is associated with every ratio depending on the quadrant that the angle is in.