Tan 30 Degrees : Exact Value, 1/√3, and How to Find It

What Does Tan 30 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 $\tan 30^\circ$ asks: in a right triangle with a 30° angle, what fraction of the adjacent side is the opposite 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 $30^\circ$ the radius meets the circle at $\left(\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}\right)$, so the slope is $\dfrac{1/2}{\sqrt{3}/2} = \dfrac{1}{\sqrt{3}}$.

Where Does Tan 30 Degrees Show Up?

A 30° incline is a gentle grade - the shallow end of a wheelchair ramp or a low roof pitch - and its slope, rise over run, is $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$, so it rises only about $0.58$ m for every metre it travels forward. That is the same short-leg-to-long-leg ratio in the 30-60-90 set square used in drafting.

In optics and physics, a 30° angle of refraction or launch feeds $\tan 30^\circ$ into the geometry, and the value connects directly to the trigonometric table of standard angles that every problem set draws from.

Standard-Angle Reference Table

Thirty 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.

Tan 30° is the smallest non-zero tangent in this set - below $1$ because $30^\circ$ is below the $45^\circ$ crossover. It is also the reciprocal of $\tan 60^\circ$: $\dfrac{1}{\sqrt{3}}$ and $\sqrt{3}$ multiply to $1$.

How Do You Find The Exact Value Of Tan 30 Degrees?

There are three clean routes, and all three land on $\dfrac{1}{\sqrt{3}}$.

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 $30^\circ$ is $1$ (half of the base that got split),

• the side adjacent to $30^\circ$ is $\sqrt{3}$, from the Pythagorean theorem: $\sqrt{2^2 - 1^2} = \sqrt{3}$.

Now apply the definition:

$$\tan 30^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$

The last step rationalises the denominator by multiplying top and bottom by $\sqrt{3}$.

Method 2: The unit-circle slope.

Tangent equals $\dfrac{\sin\theta}{\cos\theta}$. Using $\sin 30^\circ = \dfrac{1}{2}$ and $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$:

$$\tan 30^\circ = \frac{\sin 30^\circ}{\cos 30^\circ} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$$

Method 3: The reciprocal of $\tan 60^\circ$.

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

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

A calculator in degree mode confirms it: $\tan(30) = 0.5773502\ldots$, which is $\dfrac{\sqrt{3}}{3}$.

Examples Of Tan 30 Degrees

Example 1

Evaluate $6\tan 30^\circ$.

$$6\tan 30^\circ = 6 \times \frac{1}{\sqrt{3}} = \frac{6}{\sqrt{3}} = 2\sqrt{3} \approx 3.464$$

Example 2

A right triangle has a 30° angle and the side adjacent to it measures $9$ cm. Find the side opposite the 30° angle.

Wrong attempt. A student writes $\tan 30^\circ = \dfrac{\text{adjacent}}{\text{opposite}}$, giving $\dfrac{1}{\sqrt{3}} = \dfrac{9}{\text{opposite}}$ and opposite $= 9\sqrt{3} \approx 15.6$ cm.

That breaks on a quick check: the side opposite the small $30^\circ$ angle should be the shorter one, yet this answer is longer than the $9$ cm adjacent side. The ratio was flipped.

Correct. Tangent is opposite over adjacent, so $\tan 30^\circ = \dfrac{\text{opposite}}{9}$, which gives opposite $= 9 \times \dfrac{1}{\sqrt{3}} = 3\sqrt{3} \approx 5.20$ cm. Now the opposite side is shorter, as it must be.

Example 3

From a point $30$ m from the base of a tower, the angle of elevation to the top is 30°. How tall is the tower?

$$\text{height} = 30 \times \tan 30^\circ = 30 \times \frac{1}{\sqrt{3}} = 10\sqrt{3} \approx 17.3 \text{ m}$$

Example 4

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

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

The product is $1$ because the two angles are complementary and their tangents are reciprocals.

Example 5

Express $\tan 30^\circ$ in radians and evaluate $\tan\left(\dfrac{\pi}{6}\right)$.

Since $30^\circ = \dfrac{\pi}{6}$ radians, $\tan\left(\dfrac{\pi}{6}\right) = \tan 30^\circ = \dfrac{1}{\sqrt{3}}$. The twin article tan π/6 leads from the radian and unit-circle side, but the value is identical.

Where Students Trip Up On Tan 30 Degrees

Mistake 1: Swapping tan 30° and tan 60°

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

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

The correct way: The smaller angle has the smaller tangent. $\tan 30^\circ = \dfrac{1}{\sqrt{3}} \approx 0.58$; $\tan 60^\circ = \sqrt{3} \approx 1.73$. Anchoring on "tangent grows as the angle grows" keeps the gentle 30° tied to the smaller value.

Mistake 2: Leaving the denominator unrationalised when the exact form is asked

Where it slips in: Stopping at $\dfrac{1}{\sqrt{3}}$ when the marking scheme wants a rational denominator.

Don't do this: Writing only $\dfrac{1}{\sqrt{3}}$ where the standard form is expected.

The correct way: Multiply top and bottom by $\sqrt{3}$ to get $\dfrac{\sqrt{3}}{3}$, the rationalised form most textbooks treat as final. Both are equal, but $\dfrac{\sqrt{3}}{3}$ is the presentation examiners usually mark as complete.

Mistake 3: Reading the negative sibling as the same value

Where it slips in: Assuming every angle with a 30° reference gives $+\dfrac{1}{\sqrt{3}}$.

Don't do this: Writing $\tan 150^\circ = \dfrac{1}{\sqrt{3}}$ because its reference angle is $30^\circ$.

The correct way: In the second quadrant tangent is negative, so $\tan 150^\circ = -\dfrac{1}{\sqrt{3}}$. Its radian form, tan 5π/6, carries that same negative sign even though it shares the $30^\circ$ reference angle.

Key Takeaways

• Tan 30 degrees equals $\dfrac{1}{\sqrt{3}}$, or $\dfrac{\sqrt{3}}{3}$ rationalised, approximately $0.5774$ — an exact value because $30^\circ$ is a standard angle.

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

• In radians, $\tan 30^\circ = \tan\left(\dfrac{\pi}{6}\right)$, and it is the reciprocal of tan 60 degrees.

• The common slips are swapping it with $\tan 60^\circ = \sqrt{3}$ and leaving $\dfrac{1}{\sqrt{3}}$ unrationalised when $\dfrac{\sqrt{3}}{3}$ is expected.

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 $\tan 30^\circ + \tan 60^\circ$ and leave the answer with a rational denominator.

2. A ramp rises at 30° over a horizontal run of $12$ m. Use $\tan 30^\circ$ to find its vertical rise.

3. Show that $\dfrac{\tan 60^\circ}{\tan 30^\circ} = 3$.

Want a live Bhanzu trainer to walk through more tan 30 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

• Pythagorean identities and how they connect the ratios

• The unit circle and where every standard angle lives