What Is The Value Of Tan 50 Degrees?
The value of tan 50 degrees is approximately $\tan 50^\circ \approx 1.1918$, rounded to four decimal places (the fuller value is $1.191753592$). Written with the angle in radians, the same statement is $\tan\frac{5\pi}{18} \approx 1.1918$, since $50^\circ$ converts to $\frac{5\pi}{18}$ radians. The result is positive.
Here is the honest part that most pages skip. The special angles $30^\circ$, $45^\circ$, and $60^\circ$ have tidy exact values such as $\tan 45^\circ = 1$ and $\tan 60^\circ = \sqrt{3}$, because those angles can be built with a ruler and compass. The angle $50^\circ$ cannot, so $\tan 50^\circ$ is non-constructible: it has no finite expression using ordinary square roots. For real work you use its decimal, $1.1918$, or the exact cofunction identity below.
$$\tan 50^\circ \approx 1.1918 \qquad \tan 50^\circ = \cot 40^\circ$$
How Do You Find Tan 50 Degrees?
You cannot derive $\tan 50^\circ$ from the special-angle triangles, but three routes give it reliably. Each one is worth knowing, because each answers a slightly different exam question.
The ratio route (right triangle). By definition the tangent function is opposite over adjacent. In a right triangle with a $50^\circ$ angle, divide the side opposite the angle by the side adjacent to it. Using the exact ratios $\sin 50^\circ = 0.7660$ and $\cos 50^\circ = 0.6428$: $$\tan 50^\circ = \frac{\sin 50^\circ}{\cos 50^\circ} = \frac{0.7660}{0.6428} \approx 1.1918$$
The cofunction route (exact relationship). Since $50^\circ + 40^\circ = 90^\circ$, the two angles are complementary, and the tangent of one equals the cotangent of the other. This is the one clean exact form the angle has. $$\tan 50^\circ = \cot(90^\circ - 50^\circ) = \cot 40^\circ$$
The calculator or table route (the practical value). Set a calculator to degree mode and read $\tan 50 = 1.1918$, or look the value up in a printed trigonometric table. Inside the calculator, a fast algorithm (a power series or a method called CORDIC) does the arithmetic, which is how the decimal is generated in the first place.
The angle sits between $45^\circ$ and $60^\circ$, so a quick sanity check is that $\tan 50^\circ$ should land between $\tan 45^\circ = 1$ and $\tan 60^\circ \approx 1.7321$. The value $1.1918$ does, which is a good habit before you trust any answer.
Where Does 50 Degrees Sit On The Unit Circle?
On the unit circle, draw a radius at $50^\circ$ measured counter-clockwise from the positive x-axis. The point where that radius meets the circle has coordinates $(\cos 50^\circ, \sin 50^\circ) = (0.6428, 0.7660)$. Because $50^\circ$ is between $0^\circ$ and $90^\circ$, the point lands in the first quadrant, where both coordinates are positive.
The tangent is the y-coordinate divided by the x-coordinate of that point:
$$\tan 50^\circ = \frac{y}{x} = \frac{\sin 50^\circ}{\cos 50^\circ} = \frac{0.7660}{0.6428} \approx 1.1918$$
That is the same division as the right-triangle ratio, which is the point of the unit circle: it turns a triangle relationship into a value you can read off a picture for any angle, not only the ones you can draw with a compass.
Placing $\tan 50^\circ$ next to its neighbours shows how the value grows as the angle climbs toward $90^\circ$.
Table: Sine, cosine, and tangent of nearby angles, in degrees and radians (values to four decimal places).
Angle | Radians | $\sin$ | $\cos$ | $\tan$ |
|---|---|---|---|---|
$30^\circ$ | $\frac{\pi}{6}$ | $0.5000$ | $0.8660$ | $0.5774$ |
$45^\circ$ | $\frac{\pi}{4}$ | $0.7071$ | $0.7071$ | $1.0000$ |
$50^\circ$ | $\frac{5\pi}{18}$ | $0.7660$ | $0.6428$ | $1.1918$ |
$60^\circ$ | $\frac{\pi}{3}$ | $0.8660$ | $0.5000$ | $1.7321$ |
For the full grid across every standard angle, see the trigonometric table.
Why Is Tan 50 Degrees Positive?
The sign of any tangent comes from the quadrant the angle lands in, summarised by the ASTC rule (in quadrants I, II, III, IV the positive functions are All, Sine, Tangent, Cosine). The reasoning for $50^\circ$ is short.
The angle is in Quadrant I. Any angle between $0^\circ$ and $90^\circ$ sits in the first quadrant, where sine, cosine, and tangent are all positive.
Both coordinates are positive. On the unit circle the $50^\circ$ point is $(0.6428, 0.7660)$, so $\tan 50^\circ = \frac{0.7660}{0.6428}$ is a positive number divided by a positive number.
The value exceeds 1. Past $45^\circ$ the height beats the width ($\sin 50^\circ > \cos 50^\circ$), so the ratio climbs above $1$. That is why a $50^\circ$ slope rises faster than it runs.
Contrast this with $\tan 130^\circ$. That angle is in Quadrant II, where only sine stays positive, so its tangent is negative: $\tan 130^\circ \approx -1.1918$. Same digits, opposite sign, and the difference is nothing more than the quadrant.
Who Discovered The Trigonometry Behind Tan 50 Degrees?
Nobody woke up one morning and "discovered" $\tan 50^\circ$. The value is a small consequence of a much older project: building tables that give a trigonometric ratio for every angle, so that astronomers and navigators could compute without redrawing triangles each time.
Two later figures pushed the tables toward the modern form:
Aryabhata (476-550 CE, India) compiled a table of sine values (which he called jya) in his Aryabhatiya around 500 CE, spacing the angles finely enough for real astronomy.
Madhava of Sangamagrama (c. 1340-1425 CE, India) found infinite series for sine and cosine, the ancestors of the power series a calculator still uses to compute values like $\tan 50^\circ$ to many decimal places.
Where Is Tan 50 Degrees Used In The Real World?
A specific tangent value shows up wherever a fixed slope or a line of sight sits near that angle.
Roofing and construction: a roof pitched at $50^\circ$ has a rise-to-run ratio of $1.1918$, so builders multiply the horizontal span by that number to find the vertical height of the ridge.
Ramps and accessibility: the steepness of a ramp or staircase is a tangent, and checking $\tan 50^\circ \approx 1.19$ shows instantly that a $50^\circ$ ramp is far too steep for safe use, which is why real ramps use small angles.
Surveying and navigation: to find the height of a tower, a surveyor measures the angle of elevation to its top and multiplies the ground distance by the tangent of that angle.
Astronomy and optics: the apparent shift of an object against a background depends on the tangent of the viewing angle, the same tabulation Hipparchus began.
One value, read off in seconds, does work that would otherwise need a fresh triangle drawn to scale every time. That is the quiet usefulness a trigonometric table was invented to give.
What Are The Most Common Mistakes With Tan 50 Degrees?
These four errors account for most lost marks on angles like $50^\circ$, and each maps to a confusion visible in the questions students actually post online.
Leaving the calculator in radian mode.
Where it slips in:
A student types $\tan 50$ expecting $1.1918$ but the calculator is set to radians, so it returns roughly $-0.2719$, the tangent of $50$ radians.
Don't do this:
Do not trust the display before checking the angle-unit setting.
The correct way:
Switch to degree mode for $\tan 50^\circ$, or convert first: $50^\circ = \frac{5\pi}{18}$ radians, then evaluate $\tan\frac{5\pi}{18} \approx 1.1918$.
Hunting for a clean surd form.
Where it slips in:
A student assumes $\tan 50^\circ$ must equal something like $\sqrt{3}$ or a neat fraction, because $30^\circ$, $45^\circ$, and $60^\circ$ all do.
Don't do this:
Do not invent an exact radical. The angle $50^\circ$ is non-constructible, so no finite square-root expression exists.
The correct way:
Quote the decimal $\tan 50^\circ \approx 1.1918$, or the exact cofunction form $\tan 50^\circ = \cot 40^\circ$ when an exact statement is required.
Getting the sign wrong when reusing the angle.
Where it slips in:
A student computes $\tan 130^\circ$ or $\tan 230^\circ$ from the $50^\circ$ reference angle but keeps the sign positive out of habit.
Don't do this:
Do not copy the first-quadrant sign into other quadrants.
The correct way:
Apply ASTC. The reference angle is $50^\circ$ in each case, but $\tan 130^\circ \approx -1.1918$ (Quadrant II) while $\tan 230^\circ \approx +1.1918$ (Quadrant III).
Confusing the cofunction with the reciprocal.
Where it slips in:
A student reads $\tan 50^\circ = \cot 40^\circ$ and then writes $\tan 50^\circ = \cot 50^\circ$, mixing the complementary angle up with the same angle.
Don't do this:
Do not swap the angle. The cofunction pairs $50^\circ$ with its complement $40^\circ$, not with itself.
The correct way:
Keep the complement: $\tan 50^\circ = \cot 40^\circ$. The reciprocal of $\tan 50^\circ$ is $\cot 50^\circ \approx 0.8391$, a different number.
Practice Problems On Tan 50 Degrees
Use $\tan 50^\circ \approx 1.1918$ where a decimal is needed. Answers follow each problem.
Is $\tan 50^\circ$ positive or negative, and why?
(Answer: Positive, because $50^\circ$ is in Quadrant I, where tangent is positive.)Write $\tan 50^\circ$ as the cotangent of an angle smaller than $45^\circ$.
(Answer: $\tan 50^\circ = \cot 40^\circ$, since $50^\circ + 40^\circ = 90^\circ$.)Convert $50^\circ$ to radians.
(Answer: $50^\circ \times \frac{\pi}{180} = \frac{5\pi}{18} \approx 0.8727$ radians.)A ramp rises at $50^\circ$ and covers $3\text{ m}$ horizontally. How tall is it?
(Answer: height $= 3 \times \tan 50^\circ \approx 3 \times 1.1918 = 3.5754$, about $3.58\text{ m}$.)Using the reference angle, estimate $\tan 130^\circ$.
(Answer: $130^\circ$ is in Quadrant II with reference angle $50^\circ$, so $\tan 130^\circ \approx -1.1918$.)Which is larger, $\tan 50^\circ$ or $\tan 45^\circ$?
(Answer: $\tan 50^\circ \approx 1.1918$ is larger than $\tan 45^\circ = 1$.)
Where Should You Go Next After Tan 50 Degrees?
Tan 50 degrees is one entry in a much larger pattern, and a few natural doors open from here.
The tangent function. See how tangent behaves across every angle, including where it shoots to infinity near $90^\circ$.
Sin cos tan. Tie tangent back to the two ratios it is built from, with the right-triangle definitions side by side.
Cofunction identities and trigonometric ratios of complementary angles. Understand the $\tan 50^\circ = \cot 40^\circ$ bridge as a general rule.
If your child is building these foundations, a live Bhanzu trainer teaches trigonometric values starting from the unit circle and the right triangle together in the Bhanzu trigonometry program.
Was this article helpful?
Your feedback helps us write better content