Right Triangles & Trig

GE.E.1Pythagorean theorem01/04

Pythagorean theorem.

The relationship

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

Step01of 04

In a right triangle, the two short sides (legs, length $a$ and $b$ ) and the longest side (hypotenuse, length $c$ , opposite the right angle) satisfy $a^{2} + b^{2} = c^{2}$ .

The hypotenuse is always opposite the right angle.

GE.E.1 — Pythagorean theorem.

The relationship

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

Step 01 of 04

In a right triangle, the two short sides (legs, length $a$ and $b$ ) and the longest side (hypotenuse, length $c$ , opposite the right angle) satisfy $a^{2} + b^{2} = c^{2}$ .

The hypotenuse is always opposite the right angle.

Step 02 of 04

Find the hypotenuse. If $a = 3$ , $b = 4$ :

c^{2} = 3^{2} + 4^{2} = 9 + 16 = 25 \Rightarrow c = 5

The 3-4-5 triangle is the most-recognized Pythagorean triple. Memorize a few common ones (5-12-13, 8-15-17, 7-24-25) — they show up repeatedly.

Step 03 of 04

Find a missing leg. Rearrange: $a^{2} = c^{2} - b^{2}$ . With $c = 13, b = 5$ :

a^{2} = 1 3^{2} - 5^{2} = 169 - 25 = 144 \Rightarrow a = 12

Pay attention to which side is the hypotenuse: it's the LONGEST and is opposite the right angle. If you mix up $c$ and one of the legs, the algebra will give you a wrong (often imaginary) answer.

Step 04 of 04

Converse — checking if a triangle IS right. Given three side lengths, plug into $a^{2} + b^{2} = c^{2}$ with $c$ as the biggest. If equality holds, it's a right triangle; otherwise it isn't.

$a^{2} + b^{2} = c^{2}$	→right triangle
$a^{2} + b^{2} < c^{2}$	→obtuse (the angle opposite c is > 90°)
$a^{2} + b^{2} > c^{2}$	→acute

Key insight

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

only when one angle is 90° and

c

is the side opposite. The longest side is always

c

; mislabeling sides is the #1 source of wrong answers.

GE.E.2 — Special right triangles — 45-45-90 and 30-60-90.

Step 01 of 04

Two right triangles show up so often they get exact, memorizable side ratios. Knowing them lets you skip the Pythagorean computation.

45-45-90 (isosceles right)	→legs $x : x$ , hypotenuse $x 2$
30-60-90	→short leg $x$ , long leg $x 3$ , hypotenuse $2 x$

Step 02 of 04

45-45-90. Both acute angles are equal, so both legs are equal. The hypotenuse is $2$ times either leg.

Examples:

Leg = 5	→Hypotenuse = $52$
Leg = $2$	→Hypotenuse = 2
Hypotenuse = 10	→Leg = $10/ 2 = 52$

When given the hypotenuse, divide by $2$ (and rationalize if asked).

Step 03 of 04

30-60-90. Side ratios are $1 : 3 : 2$ (short leg : long leg : hypotenuse).

Identify the short leg first — it's opposite the $30°$ angle. Then everything else falls out:

short leg = $x$	→opposite 30°
long leg = $x 3$	→opposite 60°
hypotenuse = $2 x$	→opposite 90°

Examples: short leg 4 → long leg $43$ , hypotenuse 8. Hypotenuse 14 → short leg 7, long leg $73$ .

Step 04 of 04

Both come from cutting standard shapes in half.

45-45-90	→half of a square cut along its diagonal
30-60-90	→half of an equilateral triangle cut along an altitude

That's why their ratios are exact — they're consequences of the symmetries of those shapes, not approximations.

Key insight

45-45-90:

x : x : x 2

. 30-60-90:

x : x 3 : 2 x

. Identify the short leg first in 30-60-90; identify either leg first in 45-45-90. These two ratios cover a huge fraction of geometry/test problems.

GE.E.3 — Sine, cosine, tangent ratios.

Step 01 of 04

For ANY right triangle (not just the special ones), the trig ratios relate an acute angle to two of the sides. The mnemonic is SOH-CAH-TOA.

sin θ = Opposite / Hypotenuse	→SOH
cos θ = Adjacent / Hypotenuse	→CAH
tan θ = Opposite / Adjacent	→TOA

"Opposite" and "adjacent" are RELATIVE to the angle $θ$ — opposite is across from it, adjacent is the leg touching it (not the hypotenuse).

Step 02 of 04

Worked example. Right triangle with legs 3 and 4, hypotenuse 5. Find the trig ratios for the angle $θ$ opposite the side of length 3.

sin θ = \frac{3}{5} = 0.6

cos θ = \frac{4}{5} = 0.8

tan θ = \frac{3}{4} = 0.75

For the OTHER acute angle (opposite 4), opposite and adjacent swap: $sin = 4/5, cos = 3/5, tan = 4/3$ .

Step 03 of 04

Trig values for the special angles.

$sin 30° = \frac{1}{2}$	→ $cos 30° = \frac{3}{2}$
$sin 45° = \frac{2}{2}$	→ $cos 45° = \frac{2}{2}$
$sin 60° = \frac{3}{2}$	→ $cos 60° = \frac{1}{2}$
$tan 30° = \frac{3}{3}$	→ $tan 60° = 3$
$tan 45° = 1$	→

These all come from the special right triangles in the previous skill — same ratios, just renamed in trig form.

Step 04 of 04

Co-function relationship. For complementary angles $θ$ and $90° - θ$ :

sin θ = cos (90° - θ), tan θ = \frac{1}{tan ( 90° - θ )}

That's why "co-sine" is the SINE of the COMPLEMENTARY angle. The two acute angles in a right triangle are always complementary, so their sines and cosines swap.

Key insight

SOH-CAH-TOA. "Opposite" and "adjacent" depend on which acute angle you're measuring from — switch angles, swap sides. The trig ratios are dimensionless, so they only depend on the SHAPE of the triangle, not its size.

GE.E.4 — Solve right triangles — find missing sides and angles.

Step 01 of 04

"Solving" a right triangle = finding all unknown sides and angles. The Pythagorean theorem handles "all sides given except one"; trig ratios handle "an angle and one side."

2 sides given, find a side	→Pythagorean theorem
2 sides given, find an angle	→inverse trig (sin⁻¹, cos⁻¹, tan⁻¹)
1 side + 1 acute angle, find another side	→sin / cos / tan
Right angle is always 90°	→so the two acute angles sum to 90°

Step 02 of 04

Worked example — find a side. Right triangle, hypotenuse 12, one acute angle $38°$ . Find the side OPPOSITE the $38°$ .

Use sine (opposite over hypotenuse):

sin 38° = \frac{opp}{12}

opp = 12 sin 38° \approx 12 (0.6157) \approx 7.39

Step 03 of 04

Worked example — find an angle. Right triangle, opposite leg 7, hypotenuse 10. Find the angle.

sin θ = \frac{7}{10} = 0.7

θ = sin^{- 1} (0.7) \approx 44.4°

The " $sin^{- 1}$ " key on a calculator (also called arcsin) undoes sine. Same idea for cos⁻¹ and tan⁻¹.

Step 04 of 04

Picking the right ratio. Look at what you HAVE and what you WANT.

Have angle + hyp, want opp	→sin: $opp = hyp \cdot sin θ$
Have angle + hyp, want adj	→cos: $adj = hyp \cdot cos θ$
Have angle + adj, want opp	→tan: $opp = adj \cdot tan θ$
Have opp + hyp, want angle	→ $sin^{- 1} (opp / hyp)$
Have adj + hyp, want angle	→ $cos^{- 1} (adj / hyp)$
Have opp + adj, want angle	→ $tan^{- 1} (opp / adj)$

Key insight

List the labels that involve what you HAVE and what you WANT — exactly one of sin/cos/tan will use both. That's the ratio to use. To find an angle from a side ratio, use the inverse function.

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