What Is The Value Of Tan 150 Degrees?
Tan 150 degrees equals $-\dfrac{1}{\sqrt{3}}$, which rationalises to $-\dfrac{\sqrt{3}}{3}$ and works out to about $-0.5774$. Written with the angle in both units:
$$\tan 150^\circ = \tan\frac{5\pi}{6} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} \approx -0.5774$$
The angle $150^\circ$ is the same as $\dfrac{5\pi}{6}$ radians. The two forms name the identical rotation, one measured in degrees and one in the radian measure used across higher math. If radians feel unfamiliar, the page on what is a radian sets up the idea from scratch.
The value is negative, and that sign is the whole story of this angle. Everything below explains where the $\dfrac{1}{\sqrt{3}}$ comes from and why the minus sign is attached to it.
How Do You Find Tan 150 Degrees?
Finding tan 150 degrees takes two decisions: what size the answer is, and what sign it carries. The size comes from a reference angle; the sign comes from the quadrant.
The reference angle. $150^\circ$ lands in Quadrant II. Its reference angle is the gap to the $x$-axis at $180^\circ$, so $180^\circ - 150^\circ = 30^\circ$. That fixes the size of the answer as $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$.
The sign. Use the ASTC rule (often read as "All, Sin, Tan, Cos" around the quadrants, also called CAST). In Quadrant II only sine is positive, so tangent is negative there.
Put the two together:
$$\tan 150^\circ = -\tan 30^\circ = -\frac{1}{\sqrt{3}} \approx -0.5774$$
This is the fastest route, and it works for any angle once you can name its reference angle and its quadrant. For the parent value it leans on, see tan 30 degrees.
Where Does 150° Sit On The Unit Circle?
On the unit circle, an angle of $150^\circ$ points up and to the left. The point where its ray meets the circle has coordinates:
$$\left(\cos 150^\circ,; \sin 150^\circ\right) = \left(-\frac{\sqrt{3}}{2},; \frac{1}{2}\right)$$
The tangent is defined on the circle as the $y$-coordinate divided by the $x$-coordinate:
$$\tan 150^\circ = \frac{\sin 150^\circ}{\cos 150^\circ} = \frac{\tfrac{1}{2}}{-\tfrac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}}$$
The $y$-value is positive (the point is above the axis) and the $x$-value is negative (the point is to the left). A positive divided by a negative is negative, so the tangent is negative. That single sign pattern is the unit-circle reason for the minus.
For an interactive version of this picture, the unit circle with tangent walkthrough shows how the tangent value changes as the angle moves around the circle.
Can You Derive Tan 150 Degrees From Tan 30 Degrees?
Yes, and it is worth seeing the algebra rather than only the picture. Because $150^\circ = 180^\circ - 30^\circ$, the tangent difference identity gives an exact derivation. Using $\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}$ with $A = 180^\circ$ and $B = 30^\circ$:
$$\tan 150^\circ = \tan(180^\circ - 30^\circ) = \frac{\tan 180^\circ - \tan 30^\circ}{1 + \tan 180^\circ \tan 30^\circ}$$
Now substitute $\tan 180^\circ = 0$ and $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$:
$$\tan 150^\circ = \frac{0 - \tfrac{1}{\sqrt{3}}}{1 + (0)\left(\tfrac{1}{\sqrt{3}}\right)} = \frac{-\tfrac{1}{\sqrt{3}}}{1} = -\frac{1}{\sqrt{3}}$$
That collapses to the shortcut identity $\tan(180^\circ - \theta) = -\tan\theta$, which is why the reference-angle method and the algebra agree. The right-triangle side of the picture is the same $30^\circ$ triangle: a 30-60-90 triangle has legs in the ratio $1 : \sqrt{3}$, so the tangent of its $30^\circ$ angle (opposite over adjacent) is $\dfrac{1}{\sqrt{3}}$. The unit circle then supplies the minus sign the bare triangle cannot show.
What Are The Related Values Around Tan 150 Degrees?
Placing tan 150 degrees next to its neighbours makes the pattern of signs and sizes clear. Every obtuse angle below sits in Quadrant II, so each tangent is negative.
Table: Sine, cosine, and tangent for the special angles around 150°, in degrees and radians.
Angle | Radians | $\sin$ | $\cos$ | $\tan$ |
|---|---|---|---|---|
$30^\circ$ | $\tfrac{\pi}{6}$ | $\tfrac{1}{2}$ | $\tfrac{\sqrt{3}}{2}$ | $\tfrac{1}{\sqrt{3}} \approx 0.577$ |
$120^\circ$ | $\tfrac{2\pi}{3}$ | $\tfrac{\sqrt{3}}{2}$ | $-\tfrac{1}{2}$ | $-\sqrt{3} \approx -1.732$ |
$135^\circ$ | $\tfrac{3\pi}{4}$ | $\tfrac{\sqrt{2}}{2}$ | $-\tfrac{\sqrt{2}}{2}$ | $-1$ |
$150^\circ$ | $\tfrac{5\pi}{6}$ | $\tfrac{1}{2}$ | $-\tfrac{\sqrt{3}}{2}$ | $-\tfrac{1}{\sqrt{3}} \approx -0.577$ |
Notice the mirror: $\tan 30^\circ$ and $\tan 150^\circ$ have the same size and opposite signs, because $150^\circ$ is $30^\circ$ reflected across the vertical. A full grid of these values lives in the trigonometric table, and the method behind them is in trigonometric ratios of specific angles.
Why Is Tan 150 Degrees Negative?
The size $\dfrac{1}{\sqrt{3}}$ is never in doubt; the minus sign is where marks are won or lost. Three ways of looking at it all point to the same answer.
Coordinate signs. At $150^\circ$ the unit-circle point is $\left(-\tfrac{\sqrt{3}}{2}, \tfrac{1}{2}\right)$. Tangent is $\dfrac{y}{x}$, and $\dfrac{+}{-}$ is negative.
Quadrant rule. In Quadrant II sine is positive but cosine is negative, so their quotient, tangent, is negative. Only sine survives as positive there.
Slope reading. Tangent of an angle is the slope of the ray at that angle. A $150^\circ$ ray tilts downhill from left to right, and a downhill line has a negative slope.
The reference angle sets the number, and the quadrant sets the sign. Keep those two jobs separate and the answer falls out every time. The reciprocal view is worth a glance too: since tangent and cotangent are reciprocals, $\cot 150^\circ = -\sqrt{3}$, which you can confirm from the reciprocal identities.
Who Discovered The Math Behind Tan 150 Degrees?
Nobody woke up one morning and "found" tan 150 degrees. The value is the product of centuries of table-building, back when every sine and tangent had to be computed by hand and written down for sailors, astronomers, and surveyors to use.
Two other figures shaped the tables this value rests on:
Hipparchus of Nicaea (c. 190–120 BCE, Greece) built the earliest known table of chords around 150 BCE, the ancestor of the sine and tangent tables, to predict the positions of stars.
Madhava of Sangamagrama (c. 1340–1425 CE, India) found infinite series that compute sine and cosine to any accuracy, so a value like $-\dfrac{1}{\sqrt{3}}$ could be pinned down to as many decimals as anyone wanted, the same idea a calculator uses today.
Where Is Tan 150 Degrees Used In The Real World?
Obtuse angles and their negative tangents are not a classroom curiosity. A negative tangent shows up wherever a direction points "back and over" past the vertical.
Ramps and roads. A road descending from a crest makes an obtuse angle with the forward direction, and its grade, a slope, is a tangent that reads negative going down.
Navigation and bearings. A heading in the second quadrant of a compass frame produces a tangent with the sign that tells a pilot or sailor which way the track leans.
Physics of projectiles. On the way down, a projectile's velocity makes an obtuse angle with the horizontal, and the tangent of that angle is negative, encoding the descent.
Computer graphics. Rotating a sprite or camera past $90^\circ$ moves it into the second quadrant, where the engine relies on the correct negative tangent to place it.
One value, $-\dfrac{1}{\sqrt{3}}$, quietly encodes "up but leaning backward" across roads, radar, and rendering. The tangent is how the underlying tangent function turns an angle into a slope.
What Are The Most Common Mistakes With Tan 150 Degrees?
These four errors account for most lost marks on this angle, and each is surfaced in the student Q&A threads on tan(150) and tan(−150).
Dropping the negative sign.
Where it slips in:
The memorizer recalls $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$, sees the $30^\circ$ reference angle, and writes the answer as positive.
Don't do this:
Do not report $\tan 150^\circ = \dfrac{1}{\sqrt{3}}$. The reference angle gives only the size.
The correct way:
Check the quadrant before writing the sign. $150^\circ$ is in Quadrant II, tangent is negative there, so $\tan 150^\circ = -\dfrac{1}{\sqrt{3}}$.
Leaving the calculator in the wrong MODE.
Where it slips in:
The rusher types tan(150) with the calculator set to radians and reads off $\approx -6.55$, an answer for $150$ radians, not $150$ degrees.
Don't do this:
Do not trust the display until the angle unit is confirmed.
The correct way:
Set the mode to degrees for $150^\circ$, or convert to radians and enter $\dfrac{5\pi}{6}$. Either way the value is $-0.5774$.
Building the reference angle from the wrong number.
Where it slips in:
A student subtracts from $90^\circ$ and gets a reference angle of $60^\circ$, or writes $150 - 90 = 60$, misplacing the quadrant.
Don't do this:
Do not use $90^\circ$ as the anchor for a Quadrant II angle.
The correct way:
For a Quadrant II angle, take $180^\circ - \theta$. Here $180^\circ - 150^\circ = 30^\circ$, so the reference angle is $30^\circ$.
Confusing tan 150° with tan 210°.
Where it slips in:
The second-guesser assumes any angle with a $30^\circ$ reference angle gives the same tangent, so tan 150° and tan 210° get treated as equal.
Don't do this:
Do not copy the sign across quadrants. $210^\circ$ is in Quadrant III, where tangent is positive.
The correct way:
Read each quadrant separately: $\tan 150^\circ = -\dfrac{1}{\sqrt{3}}$ (Quadrant II), but $\tan 210^\circ = +\dfrac{1}{\sqrt{3}}$ (Quadrant III).
Practice Problems On Tan 150 Degrees
Work each one, then check against the answer.
Evaluate $\tan 150^\circ$ from its sine and cosine.
(Answer: $\dfrac{\sin 150^\circ}{\cos 150^\circ} = \dfrac{1/2}{-\sqrt{3}/2} = -\dfrac{1}{\sqrt{3}}$.)Convert $150^\circ$ to radians.
(Answer: $150 \times \dfrac{\pi}{180} = \dfrac{5\pi}{6}$.)State the reference angle and quadrant of $150^\circ$.
(Answer: reference angle $30^\circ$, Quadrant II.)Find $\tan 150^\circ + \tan 30^\circ$.
(Answer: $-\dfrac{1}{\sqrt{3}} + \dfrac{1}{\sqrt{3}} = 0$.)Find $\cot 150^\circ$.
(Answer: $\dfrac{1}{\tan 150^\circ} = -\sqrt{3}$.)True or false: $\tan 150^\circ = \tan 210^\circ$.
(Answer: False. $\tan 150^\circ = -\dfrac{1}{\sqrt{3}}$, but $\tan 210^\circ = +\dfrac{1}{\sqrt{3}}$.)
Where Should You Go Next After Tan 150 Degrees?
Tan 150 degrees is one doorway into how the whole unit circle behaves, and a few natural next steps open from here.
Trigonometric ratios of specific angles. Learn the reference-angle method once and every special angle becomes a two-step problem: size, then sign.
Sin, cos, tan. Go back to how the three ratios are defined together, so tangent as $\dfrac{\sin}{\cos}$ feels obvious rather than memorised.
Cofunction identities. See how angles that add to $90^\circ$ or $180^\circ$ trade values, the pattern that ties $30^\circ$ and $150^\circ$ together.
If your child is building these foundations, a live Bhanzu trainer teaches special-angle trigonometry starting from the unit circle and the "why" behind each sign in the Bhanzu trigonometry program.
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