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.
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 rightangled triangle) is a triangle in which one of the three internal angles is a right angle ($\hspace{0.2em} 90 \degree \hspace{0.2em}$).
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 –
$BC^2 + AC^2 = {\color{Red} AB} ^2$
Example
The two legs of a right triangle measure $\hspace{0.2em} 5 \text{ cm} \hspace{0.2em}$ and $\hspace{0.2em} 12 \text{ cm} \hspace{0.2em}$ in length. Find the length of the hypotenuse.
Solution
If $\hspace{0.2em} x \hspace{0.2em}$ denotes the length of the hypotenuse, according the pythagorean theorem —
$\begin{align*} x^2 \hspace{0.25em} &= \hspace{0.25em} 5^2 + 12^2 \\[1em] &= \hspace{0.25em} 169 \end{align*}$
Taking the square root on both sides, we have
$\hspace{0.25em} = \hspace{0.25em} 13$
So the hypotenuse has a length of $\hspace{0.2em} 13 \text{ cm} \hspace{0.2em}$.
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 
$\sin \theta \hspace{0.25em} = \hspace{0.25em} \frac{\text{opposite}}{\text{hypotenuse}}$

Cosine 
$\cos \theta \hspace{0.25em} = \hspace{0.25em} \frac{\text{adjacent}}{\text{hypotenuse}}$

Tangent 
$\tan \theta \hspace{0.25em} = \hspace{0.25em} \frac{\text{opposite}}{\text{adjacent}}$

Cosecant 
$\csc \theta \hspace{0.25em} = \hspace{0.25em} \frac{\text{hypotenuse}}{\text{opposite}}$

Secant 
$\sec \theta \hspace{0.25em} = \hspace{0.25em} \frac{\text{hypotenuse}}{\text{adjacent}}$

Cotangent 
$\cot \theta \hspace{0.25em} = \hspace{0.25em} \frac{\text{adjacent}}{\text{opposite}}$

As mentioned earlier, values for trigonometric ratios depend only on the reference angle $\hspace{0.2em} (\theta) \hspace{0.2em}$. This is crucial, as you'll see in the second example below.
Example
In $\hspace{0.2em} \triangle ABC \hspace{0.2em}$, $\hspace{0.2em} \angle C = 90 \degree \hspace{0.2em}$, $\hspace{0.2em} \angle B = 40 \degree \hspace{0.2em}$, and $\hspace{0.2em} AC = 5 \text{ in} \hspace{0.2em}$. Find the lengths of $\hspace{0.2em} AB \hspace{0.2em}$ and $\hspace{0.2em} BC \hspace{0.2em}$.
Solution
Let's start with a rough labeled sketch of the triangle.
Next, we know
$\sin \theta \hspace{0.25em} = \hspace{0.25em} \frac{\text{opposite}}{\text{hypotenuse}}$
So, for the given triangle
$\begin{align*} \sin B \hspace{0.25em} &= \hspace{0.25em} \frac{AC}{AB} \\[1.5em] \sin 40 \degree \hspace{0.1em} &= \hspace{0.25em} \frac{5}{AB} \end{align*}$
Now, because the value of a trigonometric ratio depends only on the angle, $\hspace{0.2em} \sin 40 \degree \hspace{0.2em}$ would be a constant. As a calculator would tell you, $\hspace{0.2em} \sin 40 \degree \hspace{0.1em} \approx \hspace{0.25em} 0.59 \hspace{0.2em}$.
Substituting the value of $\hspace{0.2em} \sin 40 \degree \hspace{0.2em}$ into the equation above, we have
$\begin{align*} 0.59 \hspace{0.25em} &\approx \hspace{0.25em} \frac{5}{BC} \\[1.5em] BC \hspace{0.25em} &\approx \hspace{0.25em} 8.47 \end{align*}$
Similarly,
$\begin{align*} \tan \theta \hspace{0.25em} &= \hspace{0.25em} \frac{\text{opposite}}{\text{adjacent}} \\[1.5em] \tan B \hspace{0.25em} &= \hspace{0.25em} \frac{AC}{BC} \\[1.5em] \tan 40 \degree \hspace{0.1em} &= \hspace{0.25em} \frac{5}{BC} \end{align*}$
Substituting $\hspace{0.2em} \tan 40 \degree \hspace{0.1em} \approx \hspace{0.25em} 0.73 \hspace{0.2em}$,
$\begin{align*} 0.73 \hspace{0.25em} &\approx \hspace{0.25em} \frac{5}{BC} \\[1.5em] BC \hspace{0.25em} &\approx \hspace{0.25em} 6.85 \end{align*}$