Right Triangle Calculator

Please provide values for any three of the six fields below. At least one of those values must be a side length.

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Please provide your input and click the calculate button
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About the Right Triangle Calculator

This right triangle calculator lets you calculate the length of the hypotenuse or a leg or the area of a right triangle. For each case, you may choose from different combinations of values to input.

Also, the calculator will give you not just the answer, but also a step by step solution. So you can use it as a great tool to learn about right triangles.

Usage Guide

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i. Valid Inputs

The calculator needs exactly two inputs, at least one of which must be a side (a leg or the hypotenuse).

All inputs can be in any of the three number formats listed below.

Note — The input for each of the two angles must be between 0\hspace{0.2em} 0 \hspace{0.2em} and 90\hspace{0.2em} 90 \hspace{0.2em}.

ii. Example

If you would like to see an example of the calculator's working, just click the "example" button.

iii. Solutions

As mentioned earlier, the calculator won't just tell you the answer but also the steps you can follow to do the calculation yourself. The "show/hide solution" button would be available to you after the calculator has processed your input.

iv. Share

We would love to see you share our calculators with your family, friends, or anyone else who might find it useful.

By checking the "include calculation" checkbox, you can share your calculation as well.

Here's a quick overview of what a right triangle is a few of the basic concepts related to it.

Right Triangles

A right triangle (or right-angled triangle) is a triangle in which one of the three internal angles is a right angle (90°\hspace{0.2em} 90 \degree \hspace{0.2em}).

The longest side in a right triangle (known as the hypotenuse) is the side opposite the right angle.

Pythagorean Theorem

Now one feature of right triangles that makes them so useful and important is that they obey the pythagorean theorem.

And according to the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides (called legs). So, for the triangle above –

BC2+AC2=AB2BC^2 + AC^2 = {\color{Red} AB} ^2
Example

The two legs of a right triangle measure 5 cm\hspace{0.2em} 5 \text{ cm} \hspace{0.2em} and 12 cm\hspace{0.2em} 12 \text{ cm} \hspace{0.2em} in length. Find the length of the hypotenuse.

Solution

If x\hspace{0.2em} x \hspace{0.2em} denotes the length of the hypotenuse, according the pythagorean theorem —

x2=52+122=169\begin{align*} x^2 \hspace{0.25em} &= \hspace{0.25em} 5^2 + 12^2 \\[1em] &= \hspace{0.25em} 169 \end{align*}

Taking the square root on both sides, we have

=13\hspace{0.25em} = \hspace{0.25em} 13

So the hypotenuse has a length of 13 cm\hspace{0.2em} 13 \text{ cm} \hspace{0.2em}.

Trigonometric Ratios

Another concept that makes right triangles great for the study of triangles is that of trigonometric ratios. Here's a brief explanation.

We'll start with the figure below.

Now, when we talk about trigonometric ratios, those ratios are with respect to a reference angle. And that reference angle can be any of the two acute angles in a right triangle.

Also, as the figure shows, we have special names for the two legs. The one opposite to the angle is termed "opposite" and the one adjacent to it is called "adjacent".

Trigonometric ratios are ratios between the side lengths of a right triangle. And the value of a trigonometric ratio depends on the reference angle alone.

Here's a table listing the six trigonometric ratios.

Ratio Formula
Sine
sin⁡θ=oppositehypotenuse\sin \theta \hspace{0.25em} = \hspace{0.25em} \frac{\text{opposite}}{\text{hypotenuse}}
Cosine
cos⁡θ=adjacenthypotenuse\cos \theta \hspace{0.25em} = \hspace{0.25em} \frac{\text{adjacent}}{\text{hypotenuse}}
Tangent
tan⁡θ=oppositeadjacent\tan \theta \hspace{0.25em} = \hspace{0.25em} \frac{\text{opposite}}{\text{adjacent}}
Cosecant
csc⁡θ=hypotenuseopposite\csc \theta \hspace{0.25em} = \hspace{0.25em} \frac{\text{hypotenuse}}{\text{opposite}}
Secant
sec⁡θ=hypotenuseadjacent\sec \theta \hspace{0.25em} = \hspace{0.25em} \frac{\text{hypotenuse}}{\text{adjacent}}
Cotangent
cot⁡θ=adjacentopposite\cot \theta \hspace{0.25em} = \hspace{0.25em} \frac{\text{adjacent}}{\text{opposite}}

As mentioned earlier, values for trigonometric ratios depend only on the reference angle (θ)\hspace{0.2em} (\theta) \hspace{0.2em}. This is crucial, as you'll see in the second example below.

Example

In △ABC\hspace{0.2em} \triangle ABC \hspace{0.2em}, ∠C=90°\hspace{0.2em} \angle C = 90 \degree \hspace{0.2em}, ∠B=40°\hspace{0.2em} \angle B = 40 \degree \hspace{0.2em}, and AC=5 in\hspace{0.2em} AC = 5 \text{ in} \hspace{0.2em}. Find the lengths of AB\hspace{0.2em} AB \hspace{0.2em} and BC\hspace{0.2em} BC \hspace{0.2em}.

Solution

Let's start with a rough labeled sketch of the triangle.

Next, we know

sin⁡θ=oppositehypotenuse\sin \theta \hspace{0.25em} = \hspace{0.25em} \frac{\text{opposite}}{\text{hypotenuse}}

So, for the given triangle

sin⁡B=ACABsin⁡40°=5AB\begin{align*} \sin B \hspace{0.25em} &= \hspace{0.25em} \frac{AC}{AB} \\[1.5em] \sin 40 \degree \hspace{0.1em} &= \hspace{0.25em} \frac{5}{AB} \end{align*}

Now, because the value of a trigonometric ratio depends only on the angle, sin⁡40°\hspace{0.2em} \sin 40 \degree \hspace{0.2em} would be a constant. As a calculator would tell you, sin⁡40°≈0.59\hspace{0.2em} \sin 40 \degree \hspace{0.1em} \approx \hspace{0.25em} 0.59 \hspace{0.2em}.

Substituting the value of sin⁡40°\hspace{0.2em} \sin 40 \degree \hspace{0.2em} into the equation above, we have

0.59≈5BCBC≈8.47\begin{align*} 0.59 \hspace{0.25em} &\approx \hspace{0.25em} \frac{5}{BC} \\[1.5em] BC \hspace{0.25em} &\approx \hspace{0.25em} 8.47 \end{align*}

Similarly,

tan⁡θ=oppositeadjacenttan⁡B=ACBCtan⁡40°=5BC\begin{align*} \tan \theta \hspace{0.25em} &= \hspace{0.25em} \frac{\text{opposite}}{\text{adjacent}} \\[1.5em] \tan B \hspace{0.25em} &= \hspace{0.25em} \frac{AC}{BC} \\[1.5em] \tan 40 \degree \hspace{0.1em} &= \hspace{0.25em} \frac{5}{BC} \end{align*}

Substituting tan⁡40°≈0.73\hspace{0.2em} \tan 40 \degree \hspace{0.1em} \approx \hspace{0.25em} 0.73 \hspace{0.2em},

0.73≈5BCBC≈6.85\begin{align*} 0.73 \hspace{0.25em} &\approx \hspace{0.25em} \frac{5}{BC} \\[1.5em] BC \hspace{0.25em} &\approx \hspace{0.25em} 6.85 \end{align*}

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