Triangles — A Complete, Plain-Language Refresher
A triangle is the simplest polygon: three straight edges linking three non-collinear points called vertices. Mathematicians label it with its vertices—e.g. △ABC—so we can talk about side a (BC), side b (AC), and side c (AB) without confusion.
1 Classifying Triangles
| By side length | Definition | Quick ID |
|---|---|---|
| Equilateral | all three sides equal | three tick marks |
| Isosceles | two sides equal | two tick marks |
| Scalene | no equal sides | no repeated ticks |
| By angle size | Definition | Iconic feature |
|---|---|---|
| Right | one angle is 90 ° | tiny square at the right angle |
| Acute | all angles < 90 ° | nothing special—just “sharp” |
| Obtuse | one angle > 90 ° | visibly “blunt” corner |
Fact: a triangle can never have two angles ≥ 90 °—that would exceed the 180 ° total.
2 Bedrock Properties & Theorems
Angle sum: A + B + C = 180 °.
Exterior angle: equal to the sum of the two non-adjacent interior angles.
Triangle inequality: a + b > c, b + c > a, c + a > b.
Pythagorean theorem (right triangles only):
a2+b2=c2(c is the hypotenuse)a^{2}+b^{2}=c^{2}\quad(\text{c is the hypotenuse})
Law of Sines (any triangle):
asinA=bsinB=csinC\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}
Law of Cosines (any triangle):
a2=b2+c2−2bccosAa^{2}=b^{2}+c^{2}-2bc\cos A
(Rotate the letters to find B or C.)
3 Worked Examples
| Problem | Solution sketch |
|---|---|
| Right-triangle side: a = 3, c = 5, find b. | 3² + b² = 5² → b² = 16 → b = 4 |
| Law of Sines: b = 2, B = 90 °, C = 45 °, find c. | 2/sin90°=c/sin45°2/\sin90° = c/\sin45° → c = √2 ≈ 1.414 |
| Law of Cosines: a = 8, b = 6, c = 10, find B. | B=cos−1((a2+c2−b2)/2ac)B=\cos^{-1}( (a^{2}+c^{2}-b^{2})/2ac ) → B ≈ 36.87 ° |
4 Area Formulas
| Given | Formula | Mini-example |
|---|---|---|
| Base & height | 12bh\tfrac12 bh | b = 5, h = 6 → Area = 15 |
| Two sides & included angle | 12absinC\tfrac12 ab\sin C | a = 9, b = 7, C = 30 ° → 15.75 |
| Three sides (Heron) | s(s−a)(s−b)(s−c), s=a+b+c2\sqrt{s(s-a)(s-b)(s-c)},\ s=\tfrac{a+b+c}{2} | a = 3, b = 4, c = 5 → 6 |
5 Medians, Inradius & Circumradius
5.1 Medians
A median joins a vertex to the midpoint of the opposite side. Formula for median m<sub>a</sub> to side a:
ma=122b2+2c2−a2m_{a}=\tfrac12\sqrt{2b^{2}+2c^{2}-a^{2}}
(Cycle letters for m<sub>b</sub>, m<sub>c</sub>.)
Example: a = 2, b = 3, c = 4 →
ma=122⋅32+2⋅42−22≈2.29m_{a}=\tfrac12\sqrt{2·3^{2}+2·4^{2}-2^{2}}\approx2.29.
5.2 Inradius (r)
Largest circle that fits inside:
r=Areas(s=a+b+c2)r=\frac{\text{Area}}{s}\quad\bigl(s=\tfrac{a+b+c}{2}\bigr)
5.3 Circumradius (R)
Radius of the circumscribed circle:
R=a2sinA=b2sinB=c2sinCR=\frac{a}{2\sin A}=\frac{b}{2\sin B}=\frac{c}{2\sin C}
6 Quick-Look Cheat Sheet
| Topic | Key equation |
|---|---|
| Angle sum | A + B + C = 180 ° |
| Pythagorean | a² + b² = c² |
| Law of Sines | a/sin A = b/sin B = c/sin C |
| Law of Cosines | a² = b² + c² − 2bc cos A |
| Base–height area | ½ bh |
| SAS area | ½ ab sin C |
| Heron | √[s(s−a)(s−b)(s−c)] |
| Median to a | ½√(2b² + 2c² − a²) |
| Inradius | Area / s |
| Circumradius | a / (2 sin A) |
7 Take-Home Messages
Three sides → infinite fun: classify, compute, inscribe and circumscribe.
Every triangle hides 180 ° inside and obeys the triangle inequality.
Two master tools—Law of Sines and Law of Cosines—unlock any missing side or angle.
Multiple area formulas let you pick whichever matches the data you actually have.
Keep this toolbox handy and every triangle problem—from textbook proofs to on-site surveying—will fold neatly into place.