What Does Tan 120 Degrees Mean?
Before the value makes sense, two words need pinning down. A quadrant is one of the four quarters the axes cut the plane into, numbered anticlockwise from the top right; the second quadrant (Q2) is the top-left quarter, covering angles from $90^\circ$ to $180^\circ$. A reference angle is the acute angle between the terminal side and the $x$-axis, the "how far from horizontal" measure that strips away the quadrant.
Tangent on the unit circle is the $y$-coordinate divided by the $x$-coordinate. At $120^\circ$, which is in Q2, the point is $\left(-\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)$: the height is positive but the horizontal reach is negative. A positive divided by a negative is negative, so $\tan 120^\circ$ comes out negative. Its size matches $\tan 60^\circ$ because $60^\circ$ is the reference angle.
Where Does Tan 120 Degrees Show Up?
A line drawn at $120^\circ$ from the positive $x$-axis points up and to the left, and its slope is $\tan 120^\circ = -\sqrt{3}$, a steep negative gradient. Any direction in the second quadrant, from a vector in physics to a bearing in navigation, carries this negative tangent.
The value also appears when an obtuse angle is broken into components. Resolving a force or velocity pointed at $120^\circ$ uses the same trigonometric ratios, and the negative sign is what keeps the horizontal component pointing the correct way. It all traces back to the unit circle, where the $120^\circ$ point has a negative $x$-coordinate.
Standard-Angle Reference Table
The table below crosses from the first quadrant into the second so the sign flip is visible. Watch tangent climb to $\sqrt{3}$ at $60^\circ$, break at $90^\circ$, and come back as negatives.
Angle (degrees) | Angle (radians) | $\tan\theta$ (exact) | $\tan\theta$ (decimal) |
|---|---|---|---|
$45^\circ$ | $\dfrac{\pi}{4}$ | $1$ | $1.0000$ |
$60^\circ$ | $\dfrac{\pi}{3}$ | $\sqrt{3}$ | $1.7321$ |
$90^\circ$ | $\dfrac{\pi}{2}$ | undefined | undefined |
$120^\circ$ | $\dfrac{2\pi}{3}$ | $-\sqrt{3}$ | $-1.7321$ |
$135^\circ$ | $\dfrac{3\pi}{4}$ | $-1$ | $-1.0000$ |
$150^\circ$ | $\dfrac{5\pi}{6}$ | $-\dfrac{1}{\sqrt{3}}$ | $-0.5774$ |
Notice $\tan 120^\circ$ and $\tan 60^\circ$ share the same size, $\sqrt{3}$, and differ only in sign. That pairing is the reference-angle relationship, and it is why tan 60 degrees is the value to memorise first.
How Do You Find The Exact Value Of Tan 120 Degrees?
Three routes, all landing on $-\sqrt{3}$.
Method 1: Reference angle plus quadrant sign.
Find the reference angle, then attach the sign the quadrant demands. These are two separate steps, and keeping them separate is what prevents errors.
Reference angle: $180^\circ - 120^\circ = 60^\circ$.
Size: $\tan 60^\circ = \sqrt{3}$.
Sign: in Q2, sine is positive and cosine is negative, so tangent (sine over cosine) is negative.
$$\tan 120^\circ = -\tan 60^\circ = -\sqrt{3}$$
Method 2: The unit circle.
The point at $120^\circ$ is $\left(-\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)$.
$$\tan 120^\circ = \frac{y}{x} = \frac{\ \frac{\sqrt{3}}{2}\ }{-\frac{1}{2}} = -\sqrt{3}$$
Method 3: The supplementary-angle identity.
Because $120^\circ = 180^\circ - 60^\circ$, the identity $\tan(180^\circ - \theta) = -\tan\theta$ applies directly:
$$\tan 120^\circ = \tan(180^\circ - 60^\circ) = -\tan 60^\circ = -\sqrt{3}$$
All three agree. In radians the same angle is $\dfrac{2\pi}{3}$, so tan 2π/3 is the radian-first article for this identical value, and the sum and difference identities are where the $\tan(180^\circ - \theta)$ rule comes from.
Examples Of Tan 120 Degrees
Example 1
Evaluate $\tan 120^\circ + \sqrt{3}$.
$$\tan 120^\circ + \sqrt{3} = -\sqrt{3} + \sqrt{3} = 0$$
Example 2
Find the exact value of $\tan 120^\circ$ using its reference angle.
Wrong attempt. A student finds the reference angle $60^\circ$, recalls $\tan 60^\circ = \sqrt{3}$, and writes $\tan 120^\circ = \sqrt{3}$.
Check that against the quadrant. $120^\circ$ is in Q2, where the $x$-coordinate is negative, so tangent must be negative. A positive answer cannot be right for a Q2 angle whose tangent the table lists as negative.
Correct. The reference angle gives the size, and the quadrant gives the sign, as two separate steps:
$$\tan 120^\circ = -\tan 60^\circ = -\sqrt{3}$$
The magnitude $\sqrt{3}$ was right; the missing minus sign was the whole error.
Example 3
A line makes an angle of $120^\circ$ with the positive $x$-axis. Find its slope.
The slope of a line is the tangent of its angle of inclination:
$$\text{slope} = \tan 120^\circ = -\sqrt{3} \approx -1.732$$
The line falls as it moves right, matching the negative slope.
Example 4
Verify that $\tan 120^\circ = \dfrac{\sin 120^\circ}{\cos 120^\circ}$.
Using $\sin 120^\circ = \dfrac{\sqrt{3}}{2}$ and $\cos 120^\circ = -\dfrac{1}{2}$:
$$\frac{\sin 120^\circ}{\cos 120^\circ} = \frac{\ \frac{\sqrt{3}}{2}\ }{-\frac{1}{2}} = -\sqrt{3} = \tan 120^\circ$$
The ratio of sine to cosine reproduces the value, sign included.
Example 5
Simplify $\tan 120^\circ \times \cot 120^\circ$.
Cotangent is the reciprocal of tangent, so the product of any tangent with its own cotangent is $1$:
$$\tan 120^\circ \times \cot 120^\circ = (-\sqrt{3}) \times \left(-\frac{1}{\sqrt{3}}\right) = 1$$
Both factors are negative, so their product is the positive $1$.
Where Students Trip Up On Tan 120 Degrees
Mistake 1: Dropping the negative sign
Where it slips in: After correctly finding the reference angle, when the quadrant sign is forgotten. Students first learning obtuse angles usually compute the size right and lose the sign.
Don't do this: Writing $\tan 120^\circ = \sqrt{3}$.
The correct way: Treat size and sign as two steps. Reference angle gives $\tan 60^\circ = \sqrt{3}$; Q2 makes tangent negative; the answer is $-\sqrt{3}$.
Mistake 2: Using the wrong reference angle
Where it slips in: Subtracting from $90^\circ$ instead of $180^\circ$ for a second-quadrant angle.
Don't do this: Writing the reference angle as $120^\circ - 90^\circ = 30^\circ$ and then using $\tan 30^\circ$.
The correct way: For a Q2 angle, the reference angle is $180^\circ - \theta$. Here $180^\circ - 120^\circ = 60^\circ$, so the size is $\tan 60^\circ = \sqrt{3}$, not $\tan 30^\circ$.
Mistake 3: Assuming tan 120° is undefined
Where it slips in: Confusing the obtuse angle $120^\circ$ with $90^\circ$, where tangent actually breaks.
Don't do this: Writing $\tan 120^\circ = $ undefined.
The correct way: Tangent is undefined only where cosine is $0$, at $90^\circ$. At $120^\circ$, cosine is $-\dfrac{1}{2}$, which is non-zero, so $\tan 120^\circ = -\sqrt{3}$ is fully defined.
Key Takeaways
Tan 120 degrees equals $-\sqrt{3}$, about $-1.7321$, an exact value because $120^\circ$ is a standard angle.
The reference angle is $180^\circ - 120^\circ = 60^\circ$, giving the size $\sqrt{3}$; the second quadrant supplies the negative sign.
On the unit circle the $120^\circ$ point is $\left(-\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)$, so $\tan 120^\circ = \dfrac{y}{x} = -\sqrt{3}$.
In radians, $\tan 120^\circ = \tan\left(\dfrac{2\pi}{3}\right)$, and the most common slip is dropping the negative sign.
To get quadrant signs right every time with a teacher, explore Bhanzu's trigonometry tutor, a high school math tutor, or flexible math classes online.
Practice These Before Moving On
Evaluate $2\tan 120^\circ + \tan 60^\circ$.
A line is inclined at $120^\circ$ to the horizontal. State its slope as an exact surd.
Show that $\tan 120^\circ \times \tan 60^\circ = -3$.
Want a live Bhanzu trainer to walk through more tan 120 degrees problems? Book a free demo class.
Read More
Was this article helpful?
Your feedback helps us write better content