Tan Pi/3 : Exact Value √3, and How to Find It

What Does Tan Pi/3 Mean?

Tangent is one of the core trigonometric ratios: in a right triangle it is the side opposite an angle divided by the side adjacent to it, and on a circle it is the $y$-coordinate over the $x$-coordinate of the terminal point. So $\tan\frac{\pi}{3}$ asks what fraction the opposite side is of the adjacent side at a $60^\circ$ angle.

The angle $\frac{\pi}{3}$ measures $60^\circ$ once you convert from radians using $180^\circ = \pi$. It lands in the first quadrant, where both coordinates are positive, so the tangent is positive.

Where Tan Pi/3 Shows Up

A $60^\circ$ incline has a slope of $\tan\frac{\pi}{3} = \sqrt{3}$, meaning it rises about $1.732$ units for every $1$ unit of horizontal run, which is why $60^\circ$ ramps and roof pitches feel steep. The same value sets the geometry of an equilateral triangle, where every angle is $60^\circ$ and the height relates to the base through $\sqrt{3}$.

Because $\frac{\pi}{3}$ equals 60 degrees, the value carries into any calculation that uses hexagons or equilateral bracing, where $60^\circ$ angles repeat and the tangent $\sqrt{3}$ describes each rise-over-run.

Standard-Angle Tangent Reference Table

One-third of $\pi$ is a first-quadrant angle, so its tangent is a clean positive value. Reading down, tangent grows from $0$ toward the break at $\frac{\pi}{2}$, and $\frac{\pi}{3}$ sits near the top of the defined range.

Notice that $\tan\frac{\pi}{3} = \sqrt{3}$ is the reciprocal of $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$. That is the cofunction link at work: $\frac{\pi}{3}$ and $\frac{\pi}{6}$ are complementary angles.

How Do You Find the Exact Value of Tan Pi/3?

Three routes all give $\sqrt{3}$.

Method 1: The 30-60-90 triangle.

Take an equilateral triangle with each side $2$ and drop a perpendicular from one vertex, splitting it into two right triangles with angles $30^\circ$, $60^\circ$, and $90^\circ$. In one of them:

• the side opposite the $60^\circ$ angle is $\sqrt{3}$,

• the side adjacent to the $60^\circ$ angle is $1$.

$$\tan\frac{\pi}{3} = \tan 60^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{3}}{1} = \sqrt{3}$$

Method 2: The unit circle.

The terminal point at $\frac{\pi}{3}$ on the unit circle is $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$.

$$\tan\frac{\pi}{3} = \frac{y}{x} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$$

Method 3: Sine over cosine.

$$\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}, \qquad \cos\frac{\pi}{3} = \frac{1}{2}$$

$$\tan\frac{\pi}{3} = \frac{\sin\frac{\pi}{3}}{\cos\frac{\pi}{3}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$$

Examples of Tan Pi/3

Working with $\frac{\pi}{3}$ means working with $60^\circ$ and the number $\sqrt{3}$. The set below builds from a plain evaluation to a triangle problem and an equation. A pattern worth naming: learners rushing through the standard angles often attach $\frac{1}{\sqrt{3}}$ to $\frac{\pi}{3}$, because the two look alike on a memorised table and only the position separates them.

Example 1

Evaluate $2\tan\frac{\pi}{3}$.

$$2\tan\frac{\pi}{3} = 2 \times \sqrt{3} = 2\sqrt{3} \approx 3.464$$

Example 2

Find $\tan\frac{\pi}{3}$ from the standard-angle table.

Wrong attempt. A student recalls the values $\frac{1}{\sqrt{3}}, 1, \sqrt{3}$ but pins the wrong one to $\frac{\pi}{3}$, writing $\tan\frac{\pi}{3} = \frac{1}{\sqrt{3}}$.

That breaks against a size check. In the first quadrant the tangent grows as the angle grows, so $\tan 60^\circ$ must be larger than $\tan 45^\circ = 1$. The value $\frac{1}{\sqrt{3}} \approx 0.577$ is smaller than $1$, so it cannot be $\tan\frac{\pi}{3}$.

Correct. The larger angle gets the larger tangent: $\tan\frac{\pi}{3} = \sqrt{3} \approx 1.732$, while $\frac{1}{\sqrt{3}}$ belongs to $\frac{\pi}{6}$.

Example 3

A right triangle has a $60^\circ$ angle and an adjacent side of $4$ cm. Find the opposite side.

$$\tan\frac{\pi}{3} = \frac{\text{opposite}}{4} \implies \text{opposite} = 4 \times \sqrt{3} = 4\sqrt{3} \approx 6.93 \text{ cm}$$

Example 4

Verify that $\tan\frac{\pi}{3} \times \tan\frac{\pi}{6} = 1$.

$$\sqrt{3} \times \frac{1}{\sqrt{3}} = 1$$

The product is $1$ because $\frac{\pi}{3}$ and $\frac{\pi}{6}$ are complementary, so their tangents are reciprocals.

Example 5

Solve $\tan\theta = \sqrt{3}$ for $\theta$ in $[0, \pi)$.

Tangent equals $\sqrt{3}$ at its reference angle $\frac{\pi}{3}$ within one period.

$$\theta = \frac{\pi}{3}$$

So $\frac{\pi}{3}$ is the principal solution of $\tan\theta = \sqrt{3}$ on this interval.

Where Do Students Trip Up on Tan Pi/3?

Mistake 1: Swapping tan pi/3 and tan pi/6

Where it slips in: Recall under time pressure, when $\sqrt{3}$ and $\frac{1}{\sqrt{3}}$ get attached to the wrong angle.

Don't do this: Writing $\tan\frac{\pi}{3} = \frac{1}{\sqrt{3}}$, which is $\tan\frac{\pi}{6}$.

The correct way: The larger first-quadrant angle has the larger tangent. $\tan\frac{\pi}{3} = \sqrt{3} \approx 1.73$; $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}} \approx 0.58$. The learner who checks the value against $\tan 45^\circ = 1$ catches the swap immediately.

Mistake 2: Reporting the decimal when an exact value is asked

Where it slips in: Calculator-first solving, where the screen reads $1.732$ and the student copies it.

Don't do this: Writing $\tan\frac{\pi}{3} = 1.732$ on a problem that asks for the exact value.

The correct way: Give the exact form $\sqrt{3}$. The decimal $1.732$ is a rounded approximation of the irrational $\sqrt{3}$.

Mistake 3: Flipping opposite and adjacent in the triangle

Where it slips in: Reading the 30-60-90 triangle from the $30^\circ$ angle instead of the $60^\circ$ angle.

Don't do this: Computing $\frac{1}{\sqrt{3}}$ by putting the short side over the long side for the $60^\circ$ angle.

The correct way: For the $60^\circ$ angle, the opposite side is $\sqrt{3}$ and the adjacent side is $1$, so $\tan\frac{\pi}{3} = \frac{\sqrt{3}}{1} = \sqrt{3}$.

Key Takeaways

• Tan pi/3 equals $\sqrt{3}$, about $1.732$, a positive first-quadrant value.

• The 30-60-90 triangle gives it as opposite over adjacent, $\frac{\sqrt{3}}{1}$; the unit circle gives it as $\frac{y}{x}$ at $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$.

• In degrees, $\tan\frac{\pi}{3} = \tan 60^\circ$, and it is the reciprocal of $\tan\frac{\pi}{6}$.

• The common slip is swapping it with $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$; the larger angle has the larger tangent.

To master the standard angles with a teacher, explore Bhanzu's trigonometry tutor sessions, a high school math tutor, or math classes online.

Practice These Before Moving On

1. Evaluate $\tan\frac{\pi}{3} - \tan\frac{\pi}{6}$.

2. A ramp rises at $60^\circ$ over a horizontal run of $5$ m. Use $\tan\frac{\pi}{3}$ to find the vertical rise.

3. Solve $\tan\theta = \sqrt{3}$ for all $\theta$ in $[0, 2\pi)$.

Want a live Bhanzu trainer to walk through more tan pi/3 problems? Book a free demo class.

Read More

• Sin, cos, and tan explained — the three ratios and how tangent connects to sine and cosine.

• Tangent formula and derivation — where $\tan\theta = \frac{\sin\theta}{\cos\theta}$ comes from.

• Trigonometric ratios of specific angles — the full standard-angle set.

• Trigonometric ratios in radians — the radian view of the same values.

• Tan pi/6 exact value — the complementary angle whose tangent is the reciprocal of tan pi/3.

• Tan 60 degrees — the same value written in degrees.