# Angle Addition Postulate: Formula & Examples TL;DR The angle addition postulate states that if a point BBB lies in the interior of ∠AOC, then the two smaller angles add to the whole: ∠AOB + ∠BOC = ∠AOC. This article covers the definition, the formula, how to use it to solve for an unknown angle and six worked examples. ## What Is the Angle Addition Postulate? The **angle addition postulate** states that if a point BBB lies in the **interior** of ∠AOC — that is, ray OBOB falls between rays OAO and OCO — then the measures of the two smaller angles add up to the measure of the larger one: ∠AOB + ∠BOC = ∠AOC. A few words carry the whole idea. The three rays share one **common vertex**, the point OOO. The middle ray, OBOB, is the **common arm** shared by both smaller angles. And BBB must sit _inside_ the big angle, not outside it — if ray OBOB swings past ray OCO, the two pieces no longer tile the original angle and the equation fails. Two angles that share a vertex and a common arm like this, with no overlap, are called **adjacent angles**. A **postulate** is a statement geometry accepts as true without proof, because it is self-evident and serves as a building block for the theorems that follow. The angle addition postulate is one of these foundational starting points, alongside its straight-line cousin, the segment addition postulate, which says the same thing for lengths: a point between two endpoints splits a segment into two pieces that add to the whole. ## The Angle Addition Postulate Formula For a point BBB interior to ∠AOC: ∠AOB + ∠BOC = ∠AOC. It runs in reverse just as well. If you know the whole angle and one piece, subtract to get the other: ∠BOC = ∠AOC − ∠AOB. And it extends to more than two pieces. If rays OBOB and ODOD both lie inside ∠AOC in order, then ∠AOB + ∠BOD + ∠DOC = ∠AOC. Two special cases make the postulate especially useful: - **Right angle.** If ∠AOC = 90°, then ∠AOB + ∠BOC = 90°, and the two pieces are complementary. - **Straight angle.** If A, O, C are collinear so ∠AOC = 180°, then ∠AOB + ∠BOC = 180°, and the two pieces form a linear pair. ## How Do You Use the Angle Addition Postulate to Solve for x? 1. **Name the whole angle and its value.** Often it is a right angle (90°) or a straight angle (180°), or it is given directly. 2. **Write the postulate** for the two (or more) pieces: ∠AOB + ∠BOC = ∠AOC. 3. **Substitute** the algebraic expressions for each piece and the value of the whole. 4. **Solve for x**, then back-substitute to find each angle if the problem asks for it. When you name an angle, use three points with the vertex in the middle — ∠AOB, not just "angle O" — because a single vertex can sit inside several different angles, and the three-letter name says exactly which one you mean. ## Examples of Angle Addition Postulate ### **Example 1 -** Point BBB lies in the interior of ∠AOC. If ∠AOB = 35° and ∠BOC = 50°, find ∠AOC. By the postulate, the two pieces add to the whole: ∠AOC = ∠AOB + ∠BOC = 35° + 50° = 85°. ### **Example 2 -** Ray OBOB lies inside the right angle ∠AOC = 90°. If ∠BOC = 32°, find ∠AOB. The correct move is subtraction, because the postulate run backwards isolates one piece: ∠AOB = ∠AOC − ∠BOC = 90° − 32° = 58°. ### **Example 3 -** Point BBB is interior to ∠AOC. The angles are ∠AOB = (2x + 10)° and ∠BOC = (3x)°, and ∠AOC = 80°. Find x and each smaller angle. Write the postulate and substitute: (2x + 10) + 3x = 80. Combine like terms: 5x + 10 = 80, so 5x = 70 and x = 14. Then ∠AOB = 2(14) + 10 = 38° and ∠BOC = 3(14) = 42°. Check: 38° + 42° = 80°. ### **Example 4 -** A, O, C are collinear, so ∠AOC is a straight angle. Ray OBOB stands between them with ∠AOB = (3x + 5)° and ∠BOC = (2x − 5)°. Find x. A straight angle is 180°, so the two pieces form a linear pair: (3x + 5) + (2x − 5) = 180. Simplify: 5x = 180, so x = 36. ### **Example 5 -** Three rays OBOB and ODOD lie inside ∠AOC = 120° in order. If ∠AOB = 40° and ∠DOC = 35°, find ∠BOD. Extend the postulate to three pieces: ∠AOB + ∠BOD + ∠DOC = ∠AOC. Substitute: 40 + ∠BOD + 35 = 120; ⇒ ∠BOD = 120 − 75 = 45°. ### **Example 6 -** Ray OBOB bisects ∠AOC, and ∠AOC = (6x − 4)° while ∠AOB = (2x + 8)°. Find ∠AOC. A bisector splits the angle into two equal halves, so ∠AOB = ∠BOC, and each half is: By the postulate ∠AOC = 2∠AOB: 6x − 4 = 2(2x + 8). Expand and solve: 6x − 4 = 4x + 16, so 2x = 20 and x = 10. Then ∠AOC = 6(10) − 4 = 56° (and each half is 28°). ## Why the Angle Addition Postulate Holds So Much Up The reason this near-obvious rule sits at the base of geometry is that _it converts a picture into arithmetic_, and almost every angle result downstream is that conversion applied once more. ## Key Takeaways - The **angle addition postulate** says that if BBB is interior to ∠AOC, then ∠AOB + ∠BOC = ∠AOC. - To solve for x, substitute the expressions into the postulate, solve, then back-substitute to report the actual angles.