Tan 70 Degrees: Value 2.7475, Methods and Examples

What Does Tan 70 Degrees Mean?

In a right triangle with a $70^\circ$ angle, tan 70 degrees is the side opposite that angle divided by the side adjacent to it. The opposite side is about $2.75$ times the adjacent side, which is why a $70^\circ$ slope looks steep.

On the unit circle (radius $1$, centred at the origin), the point at $70^\circ$ is $(\cos 70^\circ, \sin 70^\circ) \approx (0.3420, 0.9397)$. The tangent is $\frac{y}{x} \approx \frac{0.9397}{0.3420} \approx 2.7475$, the slope of the radius.

A quadrant is one of the four regions the axes cut the plane into, numbered I to IV anticlockwise. $70^\circ$ lies in Quadrant I, where sine and cosine are both positive, so tan 70 degrees is positive and no sign change is needed.

Standard-Angle Tangent Reference Table

Tangent grows slowly up to $45^\circ$ and then climbs fast toward the vertical asymptote at $90^\circ$. The values for $70^\circ$ and its neighbours show that climb.

From $60^\circ$ to $70^\circ$ the value jumps by about $1.02$, and from $70^\circ$ to $80^\circ$ by almost $2.92$. Rounding an angle near $70^\circ$ therefore moves the tangent much more than the same rounding near $20^\circ$.

The cofunction partner has its own page: tan 20 degrees is about $0.3640$, the reciprocal of this value.

Where Does Tan 70 Degrees Show Up, And Why Does It Matter?

Tan 70 degrees turns a horizontal distance into a height when the line of sight is steep. Standing $20$ m from a tower and looking up at its top at an angle of elevation of $70^\circ$, the top is $20 \times 2.7475 \approx 54.95$ m above eye level.

The same ratio read the other way gives shadows. When the sun is $70^\circ$ above the horizon, a $12$ m pole casts a shadow of $\frac{12}{\tan 70^\circ} \approx 4.37$ m, and a line or roof pitched at $70^\circ$ rises about $2.75$ units for every unit across.

How Do You Find The Value Of Tan 70 Degrees?

Four routes give $2.7475$. The first is the one to reach for in exams, because it turns $70^\circ$ into a problem about $20^\circ$.

Method 1: The Cofunction Identity

The cofunction identity is $\tan\theta = \cot(90^\circ - \theta)$. With $\theta = 70^\circ$:

$$\tan 70^\circ = \cot(90^\circ - 70^\circ) = \cot 20^\circ$$

$$\tan 70^\circ = \frac{1}{\tan 20^\circ} \approx \frac{1}{0.3640} \approx 2.7475$$

In the right triangle, the side opposite $70^\circ$ is the side adjacent to $20^\circ$, so the two ratios are reciprocals; the cofunction identities page lists all six pairs.

Method 2: Sine Over Cosine

$$\tan 70^\circ = \frac{\sin 70^\circ}{\cos 70^\circ} \approx \frac{0.9397}{0.3420} \approx 2.748$$

Rounded inputs give $2.748$; with full-precision sine and cosine the quotient is $2.7474774$. Since $\cos 70^\circ = \sin 20^\circ$, this is also $\frac{\sin 70^\circ}{\sin 20^\circ}$.

Method 3: A Short Series (No Calculator)

Work with $\cot 20^\circ$, because $20^\circ = \frac{\pi}{9} \approx 0.349066$ is a fairly small angle. For small $x$ in radians, $\cot x \approx \frac{1}{x} - \frac{x}{3}$:

$$\cot 20^\circ \approx \frac{1}{0.349066} - \frac{0.349066}{3}$$

$$\approx 2.864789 - 0.116355 = 2.748434$$

That is within about $0.001$ of the true $2.747477$. The bare first term $\frac{1}{x} = 2.8648$ is about $4$ too big, so it only works as a rough estimate; the approximation $\tan x \approx x$ applied directly to $70^\circ$ gives $1.2217$, which is not even close.

Method 4: A Calculator

In degree mode, $\tan(70) = 2.7474774195$, which rounds to $2.7475$. A handy fraction to remember is $\frac{11}{4} = 2.75$, which is high by about $0.0025$.

Is There An Exact Value Of Tan 70 Degrees?

The value is exact as $\cot 20^\circ$ or $\frac{\sin 70^\circ}{\sin 20^\circ}$, but it has no form built only from whole numbers and real square roots. That surprises students who know $\tan 30^\circ$, $\tan 45^\circ$ and $\tan 60^\circ$ by heart.

The reason shows up in the triple-angle formula for tangent, the one on the tan 3x page:

$$\tan 3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$$

With $\theta = 70^\circ$, $3\theta = 210^\circ$ and $\tan 210^\circ = \frac{\sqrt{3}}{3}$. Clearing fractions gives a cubic for $t = \tan 70^\circ$:

$$\sqrt{3},t^3 - 3t^2 - 3\sqrt{3},t + 1 = 0$$

Its three roots are $\tan 70^\circ \approx 2.7475$, $\tan 10^\circ \approx 0.1763$ and $\tan 130^\circ \approx -1.1918$, all real. A cubic with three real roots that does not factor neatly is the "casus irreducibilis": the cubic formula can still solve it, but only through cube roots of complex numbers.

Over the rationals, $\tan 70^\circ$ is a root of the degree-6 equation $3x^6 - 27x^4 + 33x^2 - 1 = 0$, as listed in the OEIS entry for this constant. Only whole-degree angles that are multiples of $3^\circ$ have real square-root forms, and $70^\circ$ is not one of them.

Can Tan 70 Degrees Be Written As A Fraction?

Not exactly. The value is irrational, so any fraction is an approximation: $\frac{11}{4} = 2.75$ is close, and $\frac{27475}{10000} = 2.7475$ matches four decimal places.

How Do You Write Tan 70 Degrees In Terms Of Other Trig Functions?

Each expression below equals $2.7475$ when evaluated. The cofunction and reciprocal forms are the ones that make identity problems collapse.

Examples Of Tan 70 Degrees

Example 1

Evaluate $\dfrac{3\tan 70^\circ}{\cot 20^\circ}$.

Since $\tan 70^\circ = \cot 20^\circ$:

$$\frac{3\tan 70^\circ}{\cot 20^\circ} = 3 \times 1 = 3$$

Example 2

Find the value of $3\tan 20^\circ \tan 45^\circ \tan 70^\circ$.

Wrong path first. A student adds the angles inside the product and writes $\tan 20^\circ \tan 70^\circ = \tan 90^\circ$, which is undefined.

That cannot be right. Both $\tan 20^\circ \approx 0.364$ and $\tan 70^\circ \approx 2.748$ are ordinary positive numbers, so their product is an ordinary number too, roughly $1$.

The correct route uses the cofunction link $\tan 70^\circ = \frac{1}{\tan 20^\circ}$:

$$\tan 20^\circ \times \tan 70^\circ = \tan 20^\circ \times \frac{1}{\tan 20^\circ} = 1$$

$$3\tan 20^\circ \tan 45^\circ \tan 70^\circ = 3 \times 1 \times 1 = 3$$

Example 3

Find the value of $\dfrac{\tan 70^\circ - \tan 20^\circ}{\tan 50^\circ}$.

Since $50^\circ = 70^\circ - 20^\circ$, apply the tan(A − B) formula:

$$\tan 50^\circ = \frac{\tan 70^\circ - \tan 20^\circ}{1 + \tan 70^\circ \tan 20^\circ}$$

From Example 2, $\tan 70^\circ \tan 20^\circ = 1$, so the denominator is $2$:

$$\tan 50^\circ = \frac{\tan 70^\circ - \tan 20^\circ}{2}$$

$$\frac{\tan 70^\circ - \tan 20^\circ}{\tan 50^\circ} = 2$$

Example 4

Prove that $\tan 70^\circ = \tan 20^\circ + 2\tan 50^\circ$.

Rearrange the last line of Example 3:

$$\tan 70^\circ - \tan 20^\circ = 2\tan 50^\circ$$

$$\tan 70^\circ = \tan 20^\circ + 2\tan 50^\circ$$

Check: $0.3640 + 2 \times 1.1918 = 2.7476$, which matches $2.7475$ up to rounding.

Example 5

Show that $\tan 1^\circ \tan 10^\circ \tan 20^\circ \tan 70^\circ \tan 80^\circ \tan 89^\circ = 1$.

Pair each angle with its complement, using $\tan(90^\circ - \theta) = \frac{1}{\tan\theta}$:

$$(\tan 1^\circ \tan 89^\circ)(\tan 10^\circ \tan 80^\circ)(\tan 20^\circ \tan 70^\circ)$$

$$= 1 \times 1 \times 1 = 1$$

The trick that looks optional here is the pairing; trying to multiply six decimals by hand is where the time goes.

Example 6

From a point $20$ m from the base of a tower, with eyes $1.5$ m above the ground, the angle of elevation of the top is $70^\circ$. Find the height of the tower.

$$\text{height above eye level} = 20 \times \tan 70^\circ \approx 20 \times 2.7475 = 54.95 \text{ m}$$

$$\text{tower height} \approx 54.95 + 1.5 = 56.45 \text{ m}$$

These problems are a staple of heights and distances, and the eye-height step is the one most often forgotten.

Where Do Students Trip Up On Tan 70 Degrees?

Mistake 1: A Calculator In Radian Mode

Where it slips in: Typing $\tan(70)$ into a calculator or search bar that works in radians.

Don't do this: Copying $1.2220$. That is the tangent of $70$ radians (about $4010.7^\circ$, which behaves like $50.7^\circ$ for tangent).

The correct way: Use degree mode, or enter $\tan\left(\frac{7\pi}{18}\right)$ in radians. Any angle above $45^\circ$ in Quadrant I has a tangent above $1$, and an angle this close to $90^\circ$ should give a value well above $2$.

Mistake 2: Swapping Tan 70 And Tan 20

Where it slips in: Height and shadow problems, where both the angle and its complement appear in the same diagram.

Don't do this: Writing $\tan 70^\circ \approx 0.3640$, which is $\tan 20^\circ$.

The correct way: Decide which side is opposite the angle you are using before reaching for a value. The swap between $2.7475$ and $0.3640$ usually happens when a student labels the triangle from the wrong corner, and drawing the angle on the diagram first fixes it.

Mistake 3: Using The Small-Angle Shortcut On A Large Angle

Where it slips in: Students who learned $\tan x \approx x$ for small angles apply it everywhere.

Don't do this: Writing $\tan 70^\circ \approx \frac{7\pi}{18} \approx 1.22$.

The correct way: The shortcut only works for small angles in radians, so use it on the complement instead: $\cot 20^\circ \approx \frac{1}{x} - \frac{x}{3}$, with an error near $0.001$. The memorizer who learned the shortcut without its range is the student most likely to trust the $1.22$.

Where Can You Get Help With Trigonometry Values Like Tan 70 Degrees?

Non-special angles such as $70^\circ$ are a calculator and identity skill rather than a value to memorise, and the cofunction tricks in this page are where a second explanation helps most. A trigonometry tutor, a high school math tutor or live math classes online can work through these identity problems step by step.

Key Takeaways

• Tan 70 degrees is about $2.7475$ and equals $\cot 20^\circ$ exactly.

• In radians the angle is $\frac{7\pi}{18}$, in Quadrant I, so the value is positive and above $1$.

• It has no real surd form; it solves $\sqrt{3},t^3 - 3t^2 - 3\sqrt{3},t + 1 = 0$.

• $\tan 20^\circ \tan 70^\circ = 1$ drives most exam identities, including $\tan 70^\circ = \tan 20^\circ + 2\tan 50^\circ$.

• A calculator answer of $1.2220$ means radian mode.

Practice These Before Moving On

1. Evaluate $\dfrac{8\tan 70^\circ}{9\tan 110^\circ}$.

2. Find $\sec^2 70^\circ - 1$ to four decimal places.

3. Simplify $\tan 70^\circ \times \cot 70^\circ + \tan 20^\circ \tan 70^\circ$.

4. When the sun is $70^\circ$ above the horizon, how long is the shadow of a $12$ m pole?

5. Is $\tan 70^\circ$ closer to $\frac{11}{4}$ or to $\frac{8}{3}$?

Answers:

1. $\tan 110^\circ = -\tan 70^\circ$, so the value is $-\frac{8}{9}$.

2. $\sec^2 70^\circ - 1 = \tan^2 70^\circ \approx 7.5486$.

3. $1 + 1 = 2$.

4. $\frac{12}{\tan 70^\circ} \approx 4.37$ m.

5. $\frac{11}{4} = 2.75$ is off by $0.0025$; $\frac{8}{3} \approx 2.6667$ is off by $0.0808$. So $\frac{11}{4}$ is closer.

If question 3 feels stuck, return to Example 2 and the reciprocal pairing.

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