Tan 60 Degrees : Exact Value, √3, and How to Find It

What Does Tan 60 Degrees Mean?

Tangent is one of the three core trigonometric ratios - in a right triangle, the tangent of an angle is the side opposite it divided by the side adjacent to it. So $\tan 60^\circ$ asks: in a right triangle with a 60° angle, how many times longer is the opposite side than the adjacent side?

On the unit circle - a circle of radius $1$ centred at the origin - the tangent of an angle is the slope of the radius drawn to that angle, or equivalently $\dfrac{\sin\theta}{\cos\theta}$. At $60^\circ$ the radius meets the circle at $\left(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\right)$, so the slope, and therefore the tangent, is $\dfrac{\sqrt{3}/2}{1/2} = \sqrt{3}$.

Where Does Tan 60 Degrees Show Up?

A 60° face is the pitch of the long side of a drafting set square - the 30-60-90 triangle every technical-drawing kit carries - where the long leg measures $\sqrt{3}$ times the short leg, and that ratio is $\tan 60^\circ$. The same value fixes the height of an equilateral truss: split it down the middle and the peak sits $\sqrt{3}$ units above each half-base.

In physics, a projectile launched at 60° leaves the ground with a vertical-to-horizontal speed ratio of $\tan 60^\circ = \sqrt{3}$, so it climbs far more steeply than it travels forward. The exact value sits on the unit circle, the home reference for every standard angle.

Standard-Angle Reference Table

Sixty degrees is one of a handful of angles whose tangent has a clean exact form. Here are the standard first-quadrant angles in both degrees and radians.

Read the column top to bottom and tangent climbs from $0$ upward, shooting off to infinity as the angle nears $90^\circ$ - unlike sine and cosine, it has no ceiling. Tan 60° and $\tan 30^\circ$ are reciprocals: $\tan 60^\circ = \dfrac{1}{\tan 30^\circ}$.

How Do You Find The Exact Value Of Tan 60 Degrees?

There are three clean routes, and all three land on $\sqrt{3}$.

Method 1: The 30-60-90 triangle.

Take an equilateral triangle with each side $2$ units and drop a perpendicular from one vertex to the opposite side. That splits it into two identical right triangles, each with angles $30^\circ$, $60^\circ$, and $90^\circ$.

In one of those right triangles:

• the side opposite $60^\circ$ is $\sqrt{3}$, from the Pythagorean theorem: $\sqrt{2^2 - 1^2} = \sqrt{3}$,

• the side adjacent to $60^\circ$ is $1$ (half of the base that got split).

Now apply the definition:

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

Method 2: The unit-circle slope.

Tangent equals $\dfrac{\sin\theta}{\cos\theta}$. Using the standard values $\sin 60^\circ = \dfrac{\sqrt{3}}{2}$ and $\cos 60^\circ = \dfrac{1}{2}$:

$$\tan 60^\circ = \frac{\sin 60^\circ}{\cos 60^\circ} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$$

Method 3: The reciprocal of $\tan 30^\circ$.

Because $60^\circ$ and $30^\circ$ are complementary, their tangents are reciprocals. Since $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$:

$$\tan 60^\circ = \frac{1}{\tan 30^\circ} = \frac{1}{1/\sqrt{3}} = \sqrt{3}$$

A calculator in degree mode confirms it: $\tan(60) = 1.7320508\ldots$, which is $\sqrt{3}$.

Examples Of Tan 60 Degrees

Example 1

Evaluate $5\tan 60^\circ$.

$$5\tan 60^\circ = 5 \times \sqrt{3} = 5\sqrt{3} \approx 8.660$$

Example 2

A right triangle has a 60° angle and the side adjacent to it measures $4$ cm. Find the length of the side opposite the 60° angle.

Wrong attempt. A student writes $\tan 60^\circ = \dfrac{\text{adjacent}}{\text{opposite}}$, giving $\sqrt{3} = \dfrac{4}{\text{opposite}}$ and opposite $= \dfrac{4}{\sqrt{3}} \approx 2.31$ cm.

That breaks on a quick check: the side opposite the larger angle should be the longer one, yet this answer is shorter than the adjacent side of $4$ cm. The ratio was flipped.

Correct. Tangent is opposite over adjacent, so $\tan 60^\circ = \dfrac{\text{opposite}}{4}$, which gives opposite $= 4\sqrt{3} \approx 6.93$ cm. Now the opposite side is longer, as it must be.

Example 3

A surveyor stands $20$ m from the base of a tree and measures the angle to the top as 60°. How tall is the tree?

$$\text{height} = 20 \times \tan 60^\circ = 20\sqrt{3} \approx 34.6 \text{ m}$$

Example 4

Show that $\tan 60^\circ = \dfrac{\sin 60^\circ}{\cos 60^\circ}$.

$$\frac{\sin 60^\circ}{\cos 60^\circ} = \frac{\sqrt{3}/2}{1/2} = \frac{\sqrt{3}}{2} \times \frac{2}{1} = \sqrt{3}$$

The quotient identity holds, as it does for every angle where cosine is non-zero.

Example 5

Express $\tan 60^\circ$ in radians and evaluate $\tan\left(\dfrac{\pi}{3}\right)$.

Since $60^\circ = \dfrac{\pi}{3}$ radians, $\tan\left(\dfrac{\pi}{3}\right) = \tan 60^\circ = \sqrt{3}$. The radian form and the degree form name the same angle and the same value — the twin article tan π/3 leads with the radian framing.

Where Students Trip Up On Tan 60 Degrees

Mistake 1: Swapping tan 60° and tan 30°

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

Don't do this: Writing $\tan 60^\circ = \dfrac{1}{\sqrt{3}}$. That is $\tan 30^\circ$, not $\tan 60^\circ$.

The correct way: The larger angle has the larger tangent. $\tan 60^\circ = \sqrt{3} \approx 1.73$; $\tan 30^\circ = \dfrac{1}{\sqrt{3}} \approx 0.58$. The habit that fixes this is anchoring on "tangent grows as the angle grows," so the steeper 60° must give the bigger number. The same mix-up shows up with its Q2 relative tan 2π/3, which shares the reference angle but carries a negative sign.

Mistake 2: Leaving the calculator in radian mode

Where it slips in: Calculator-first solving, where the mode was never checked.

Don't do this: Trusting a screen that reads $\tan(60) \approx 0.320$ and copying it.

The correct way: Confirm degree mode before entering $\tan(60)$; in radian mode the calculator reads $60$ as $60$ radians, not $60^\circ$. The first question to ask when a tangent value looks nothing like $1.73$ is always "what mode is the calculator in?"

Mistake 3: Assuming tangent maxes out at 1

Where it slips in: Carrying over the intuition from sine and cosine, which never exceed $1$.

Don't do this: Rejecting $\tan 60^\circ = 1.73$ because it is bigger than $1$.

The correct way: Tangent has no upper bound. It passes $1$ at $45^\circ$ and races toward infinity as the angle approaches $90^\circ$, so a value above $1$ is expected for any angle past $45^\circ$.

Key Takeaways

• Tan 60 degrees equals $\sqrt{3}$, approximately $1.7321$ — an exact value because $60^\circ$ is a standard angle.

• The 30-60-90 triangle gives it as opposite over adjacent, $\dfrac{\sqrt{3}}{1}$; the unit circle gives it as $\dfrac{\sin 60^\circ}{\cos 60^\circ}$.

• In radians, $\tan 60^\circ = \tan\left(\dfrac{\pi}{3}\right)$, and it is the reciprocal of tan 30 degrees.

• The most common slip is confusing it with $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$ — remember tangent grows as the angle grows, with no ceiling.

To take these standard angles further with a teacher, explore Bhanzu's trigonometry tutor, our high school math tutor sessions, or structured math classes online.

Practice These Before Moving On

1. Evaluate $2\tan 60^\circ - \tan 45^\circ$.

2. A ladder leans against a wall at 60° to the ground, with its foot $1.5$ m from the wall. Use $\tan 60^\circ$ to find how high up the wall it reaches.

3. Show that $\tan 60^\circ \times \tan 30^\circ = 1$.

Want a live Bhanzu trainer to walk through more tan 60 degrees problems? Book a free demo class — online, worldwide.

Read More

• Sin, cos, and tan — the three core trigonometric ratios

• Trigonometric ratios of specific angles

• The full trigonometric table of standard values

• Pythagorean identities and how they connect the ratios

• Converting 60 degrees to radians