Right Triangles & Trig

In any right triangle, the two shorter sides are the legs ($a$ and $b$) and the longest side — always across from the right angle — is the hypotenuse ($c$). The theorem says:

The two legs squared and added always equal the hypotenuse squared. Because $c$ is the biggest side, it sits alone on its own side of the equation.

Take the three sides of a triangle and let $c$ be the longest. Compare $a^2+b^2$ with $c^2$:

• If $a^2 + b^2 = c^2$, the triangle is right.
• If $a^2 + b^2 > c^2$, the triangle is acute (all angles under $90^\circ$).
• If $a^2 + b^2 < c^2$, the triangle is obtuse (one angle over $90^\circ$).

A bigger $c^2$ “pulls” the longest side apart into an obtuse angle.

A Pythagorean triple is a set of three whole numbers that fit $a^2+b^2=c^2$ exactly. Memorizing a few lets you skip the arithmetic. The common ones are:

Any multiple of a triple is also a triple. Multiply $3$-$4$-$5$ by $2$ to get $6$-$8$-$10$; by $3$ to get $9$-$12$-$15$; by $10$ to get $30$-$40$-$50$.

This is an isosceles right triangle: the two legs are equal, and each base angle is $45^\circ$. The hypotenuse is always the leg times $\sqrt{2}$:

Going up from a leg, multiply by $\sqrt{2}$. Going down from the hypotenuse, divide by $\sqrt{2}$.

The sides are always in the ratio

The short leg (across from $30^\circ$) is the key. The hypotenuse (across from $90^\circ$) is twice the short leg, and the long leg (across from $60^\circ$) is the short leg times $\sqrt3$.

Pick one of the two acute angles and call it $\theta$. Relative to $\theta$, name the sides:

• opposite — the leg across from $\theta$
• adjacent — the leg touching $\theta$ (that is not the hypotenuse)
• hypotenuse — across from the right angle

Remember it as SOH-CAH-TOA.

When you know one angle and one side and want another side, pick the ratio that uses your two sides (SOH, CAH, or TOA), then solve for the unknown. Your calculator must be in degree mode.

When you know two sides but want the angle, use the inverse functions $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$ (often the “$2$nd” + sin/cos/tan keys):

The inverse “undoes” the ratio and gives back the angle in degrees.

An angle of elevation is measured up from the horizontal to an object above you. An angle of depression is measured down from the horizontal to an object below you. Because the two horizontal lines are parallel, the angle of elevation from the ground equals the angle of depression from above (alternate interior angles).

The ratio is not memorized magic — it follows from the Pythagorean Theorem. Start with an isosceles right triangle whose two equal legs are each $x$. The hypotenuse $c$ obeys

Because both legs are equal, the two acute angles are equal, and since they must sum to $90^\circ$ each is $45^\circ$. That is exactly the pattern $x$, $x$, $x\sqrt2$.

Take an equilateral triangle with every side $2x$ and every angle $60^\circ$. Drop an altitude from the top vertex: it splits the base into two halves of length $x$ and cuts the top angle into two $30^\circ$ pieces, making a right triangle. The short leg (opposite $30^\circ$) is $x$ and the hypotenuse is $2x$, so the altitude (the long leg, opposite $60^\circ$) is

That gives the pattern $x$, $x\sqrt3$, $2x$.

In a right triangle with hypotenuse $c$, opposite side $a$, and adjacent side $b$, we have $\sin\theta = a/c$ and $\cos\theta = b/c$. Squaring and adding:

because $a^2+b^2=c^2$. So this famous Pythagorean identity is just the Pythagorean Theorem divided by $c^2$. (Here $\sin^2\theta$ means $(\sin\theta)^2$.) It holds for every angle, so if you know one of $\sin\theta$ or $\cos\theta$ you can find the other.

SOH-CAH-TOA only works in right triangles. For any triangle with angles $A$, $B$, $C$ opposite sides $a$, $b$, $c$:

The Law of Sines pairs each angle with its opposite side (use it when you know an angle-side pair). The Law of Cosines is the Pythagorean Theorem with a correction term $-2ab\cos C$; when $C = 90^\circ$, $\cos C = 0$ and it collapses back to $c^2 = a^2+b^2$.

A rectangular box with length $\ell$, width $w$, and height $h$ has a space diagonal $d$ running corner to opposite corner. Use the Pythagorean Theorem twice: first the diagonal of the base is $\sqrt{\ell^2 + w^2}$, then that base diagonal and the height $h$ form a second right triangle:

The one clean formula just adds the squares of all three dimensions.

Drop the altitude from the right angle to the hypotenuse. It splits the hypotenuse into two pieces $p$ and $q$ and creates two smaller triangles, each similar to the original. Similarity gives three “geometric mean” relations:

The altitude is the geometric mean of the two hypotenuse pieces; each leg is the geometric mean of the whole hypotenuse and the piece next to it.