What are the main right triangle relationships?

They include the Pythagorean theorem (a^2 = b^2 + c^2), the geometric-mean (leg-projection) relations (b^2 = a · b'), the altitude relations (h^2 = b' · c' and a · h = b · c), the reciprocal relation 1/(h^2) = 1/(b^2) + 1/(c^2), and the four trig ratios sine, cosine, tangent, and cotangent of an acute angle.

How do you find the altitude to the hypotenuse of a right triangle?

There are several ways: use h^2 = b' · c' (the product of the two projections), use a · h = b · c to get h = (b · c)/a, or use 1/(h^2) = 1/(b^2) + 1/(c^2). Pick whichever formula matches the values you are given.

It is a mnemonic for the three basic trig ratios: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent. It helps you set up ratios quickly for an acute angle in a right triangle.

When can you use the Pythagorean theorem?

Only for right triangles: the square of the hypotenuse equals the sum of the squares of the two legs (a^2 = b^2 + c^2). For a general (non-right) triangle, use the law of cosines instead.

Right Triangle Trigonometry: Formulas and Examples

Right triangle relationships are a cornerstone of geometry and trigonometry, and they show up constantly in coordinate geometry, physics, and standardized tests. This guide brings together the Pythagorean theorem, the geometric-mean (altitude) relations, and the SOH-CAH-TOA trig ratios, with a fully worked example that finds every side and the altitude of a right triangle.

Notation and setup

Consider triangle $A B C$ with the right angle at $A$ , and let $A H$ be the altitude drawn from $A$ down to the hypotenuse $B C$ . We name the segments as follows:

$a = B C$ is the hypotenuse;
$b = A C$ and $c = A B$ are the two legs;
$h = A H$ is the altitude to the hypotenuse;
$b^{'} = C H$ is the projection of leg $b$ onto the hypotenuse, and $c^{'} = B H$ is the projection of leg $c$ onto the hypotenuse.

Because $H$ lies between $B$ and $C$ , the two projections always add up to the hypotenuse: $b^{'} + c^{'} = a$ .

The Pythagorean theorem

In a right triangle, the square of the hypotenuse equals the sum of the squares of the two legs:

$a^{2} = b^{2} + c^{2}$

This is the most important relationship, and it lets you find any side once you know the other two. For example, a right triangle with legs $3$ and $4$ has hypotenuse $3^{2} + 4^{2} = 25 = 5$ .

Leg and projection relations

Each leg is the geometric mean of the hypotenuse and its own projection onto the hypotenuse:

$b^{2} = a \cdot b^{'} c^{2} = a \cdot c^{'}$

Adding these two relations recovers the Pythagorean theorem exactly:

$b^{2} + c^{2} = a \cdot b^{'} + a \cdot c^{'} = a (b^{'} + c^{'}) = a \cdot a = a^{2}$

Altitude relations

The altitude $h$ is tied to the remaining segments through three commonly used relations:

Relation	Meaning
$h^{2} = b^{'} \cdot c^{'}$	The altitude is the geometric mean of the two projections
$a \cdot h = b \cdot c$	Two equal ways of writing twice the area
$\frac{1}{h ^{2}} = \frac{1}{b ^{2}} + \frac{1}{c ^{2}}$	Links the altitude to the two legs

From $a \cdot h = b \cdot c$ you immediately get the formula for the altitude when all three sides are known:

$h = \frac{b \cdot c}{a}$

Trig ratios of an acute angle

For an acute angle $α$ in a right triangle, the four trig ratios are defined using the side opposite the angle, the side adjacent to it, and the hypotenuse:

$sin α = \frac{opposite}{hypotenuse} cos α = \frac{adjacent}{hypotenuse}$

$tan α = \frac{opposite}{adjacent} cot α = \frac{adjacent}{opposite}$

The first three are captured by the mnemonic SOH-CAH-TOA. A few properties and special values worth memorizing:

$tan α = \frac{sin α}{cos α}$ and $cot α = \frac{cos α}{sin α}$ , so $tan α \cdot cot α = 1$ .
$sin^{2} α + cos^{2} α = 1$ (the Pythagorean identity, true for every angle $α$ ).
If two acute angles are complementary ( $α + β = 9 0^{\circ}$ ), then $sin α = cos β$ and $tan α = cot β$ .

Angle	$3 0^{\circ}$	$4 5^{\circ}$	$6 0^{\circ}$
$sin$	$\frac{1}{2}$	$\frac{2}{2}$	$\frac{3}{2}$
$cos$	$\frac{3}{2}$	$\frac{2}{2}$	$\frac{1}{2}$
$tan$	$\frac{3}{3}$	$1$	$3$

Worked example

Problem. In triangle $A B C$ with the right angle at $A$ , the legs are $A B = 6$ cm and $A C = 8$ cm. Find the hypotenuse $B C$ , the altitude $A H$ , and the projections $B H$ and $C H$ .

Step 1 — Find the hypotenuse with the Pythagorean theorem.

$B C = A B^{2} + A C^{2} = 6^{2} + 8^{2} = 36 + 64 = 100 = 10 cm$

Step 2 — Find the altitude. Use the relation $A B \cdot A C = B C \cdot A H$ :

$A H = \frac{A B \cdot A C}{B C} = \frac{6 \cdot 8}{10} = \frac{48}{10} = 4.8 cm$

Step 3 — Find the projections. For leg $c = A B$ , its projection is $c^{'} = B H$ , so use $A B^{2} = B C \cdot B H$ :

$B H = \frac{A B ^{2}}{B C} = \frac{6 ^{2}}{10} = \frac{36}{10} = 3.6 cm$

$C H = B C - B H = 10 - 3.6 = 6.4 cm$

Check with the altitude relation $A H^{2} = B H \cdot C H$ :

$B H \cdot C H = 3.6 \cdot 6.4 = 23.04 = 4. 8^{2} = A H^{2} ✓$

The numbers match, which confirms the solution is correct.

Common mistakes

Confusing the hypotenuse with a leg. The hypotenuse is always the side opposite the right angle and the longest side — don't swap it in when applying the Pythagorean theorem.
Using the Pythagorean theorem on a non-right triangle. It holds only for right triangles; for a general triangle you must use the law of cosines.
Mixing up "opposite" and "adjacent." These are defined relative to a specific angle — change the reference angle and the opposite and adjacent sides switch.
Forgetting that $b^{'} + c^{'} = a$ . Many students compute each projection separately and forget that they must sum to the hypotenuse, skipping an easy way to check their work.

Wrap up

Master the Pythagorean theorem together with the four groups of side-and-altitude relations, and you can recover every remaining part of a right triangle from just a few starting values. Enter the numbers from the example into the calculator to check each step of the solution for yourself.