AMC 10 · 2007 · #4
Grade 8 geometry-2dThe point O is the center of the circle circumscribed about △ABC, with ∠BOC=120∘ and ∠AOB=140∘, as shown. What is the degree measure of ∠ABC?
Pick an answer.
AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: $O$ is the center of the circle that passes through all three vertices of $\triangle ABC$. Two of the central angles are known: $\angle BOC=120^\circ$ and $\angle AOB=140^\circ$. Find the measure of the triangle's angle at $B$, $\angle ABC$.
Givens: $O$ is the circumcenter of $\triangle ABC$, so $OA$, $OB$, and $OC$ are all radii of the same circle; $\angle BOC=120^\circ$; $\angle AOB=140^\circ$; Answer choices: (A) $35$, (B) $40$, (C) $45$, (D) $50$, (E) $60$ degrees
Unknowns: The degree measure of $\angle ABC$, the interior angle of the triangle at vertex $B$
Understand
Restated: $O$ is the center of the circle that passes through all three vertices of $\triangle ABC$. Two of the central angles are known: $\angle BOC=120^\circ$ and $\angle AOB=140^\circ$. Find the measure of the triangle's angle at $B$, $\angle ABC$.
Givens: $O$ is the circumcenter of $\triangle ABC$, so $OA$, $OB$, and $OC$ are all radii of the same circle; $\angle BOC=120^\circ$; $\angle AOB=140^\circ$; Answer choices: (A) $35$, (B) $40$, (C) $45$, (D) $50$, (E) $60$ degrees
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
The center $O$ is joined to all three vertices, so the picture is really three triangles $\triangle AOB$, $\triangle BOC$, $\triangle AOC$ that share the point $O$. Tool #1 (Draw a Diagram) matters because the key fact is only visible once you mark that $OA$, $OB$, $OC$ are equal radii — that makes each of the three triangles isosceles. Tool #7 (Identify Subproblems) then handles the angle at $B$ by splitting it with $OB$ into two base angles, one from each neighbouring isosceles triangle, which we can find separately and add.
Execute — Answer: D
4.MD.C.7 Step 1 Find the third central angle
- The three central angles $\angle AOB$, $\angle BOC$, and $\angle AOC$ share the vertex $O$ and together sweep once all the way around the point, so they add up to $360^\circ$.
- Subtract the two known ones to get the third: $\angle AOC=360^\circ-140^\circ-120^\circ=100^\circ$.
💡 Angles that fan out from one point and cover a full turn always total $360^\circ$.
8.G.A.5 Step 2 Base angles of triangle AOB
- Because $O$ is the circumcenter, $OA=OB$ (both radii), so $\triangle AOB$ is isosceles with its apex at $O$.
- The two base angles, at $A$ and at $B$, are equal.
- The three angles of the triangle add to $180^\circ$, so each base angle is $\dfrac{180^\circ-140^\circ}{2}=20^\circ$.
- In particular $\angle OBA=20^\circ$.
💡 Two equal radii make an isosceles triangle, and its two base angles must match.
8.G.A.5 Step 3 Base angles of triangle BOC
- The same reasoning works for $\triangle BOC$: $OB=OC$ are radii, so this triangle is isosceles too, with apex angle $\angle BOC=120^\circ$.
- Each base angle is $\dfrac{180^\circ-120^\circ}{2}=30^\circ$, so $\angle OBC=30^\circ$.
💡 A wider apex angle leaves less for the base angles, so a $120^\circ$ top gives $30^\circ$ each.
7.G.B.5 Step 4 Add the two pieces at B
- Segment $OB$ lies inside $\angle ABC$ and cuts it into the two base angles we just found: $\angle OBA$ from $\triangle AOB$ and $\angle OBC$ from $\triangle BOC$.
- Adding them gives $\angle ABC=20^\circ+30^\circ=50^\circ$.
- That is choice (D).
- (Notice $\angle AOC=100^\circ$ never had to be used here, but it confirms the picture is consistent.)
💡 The whole angle at $B$ is just its two neighbouring parts added together.
4.MD.C.7 The three central angles $\angle AOB$, $\angle BOC$, and $\angle AOC$ share the 8.G.A.5 Because $O$ is the circumcenter, $OA=OB$ (both radii), so $\triangle AOB$ is iso 8.G.A.5 The same reasoning works for $\triangle BOC$: $OB=OC$ are radii, so this triangl 7.G.B.5 Segment $OB$ lies inside $\angle ABC$ and cuts it into the two base angles we ju Review
Reasonableness: Each base angle came from a valid isosceles triangle: $20^\circ+20^\circ+140^\circ=180^\circ$ and $30^\circ+30^\circ+120^\circ=180^\circ$, both check out. The answer $50^\circ$ is a plausible triangle angle (between $0^\circ$ and $180^\circ$) and matches choice (D). As a full consistency test, the other two triangle angles come out as $\angle BAC=\tfrac{180-100}{2}+20$-type sums; more simply, each interior angle of $\triangle ABC$ is half of the central angle across from it, giving $\tfrac{120}{2}=60^\circ$ at $A$, $\tfrac{100}{2}=50^\circ$ at $B$, $\tfrac{140}{2}=70^\circ$ at $C$, and $60+50+70=180^\circ$.
Alternative: Use the inscribed-angle theorem directly. $\angle ABC$ is the inscribed angle standing on arc $AC$, and that arc is cut off by the central angle $\angle AOC=100^\circ$. An inscribed angle is half its central angle on the same arc, so $\angle ABC=\tfrac12\cdot100^\circ=50^\circ$ in one step. The isosceles-triangle method above is really the reason that half-the-central-angle rule is true.
CCSS standards used (min grade 8)
4.MD.C.7Recognize angle measure as additive (Adding the three central angles at $O$ to a full turn of $360^\circ$ to find $\angle AOC=100^\circ$.)8.G.A.5Use informal arguments to establish facts about the angle sum of triangles (Applying the $180^\circ$ angle sum inside the isosceles triangles $\triangle AOB$ and $\triangle BOC$ to get the base angles $20^\circ$ and $30^\circ$.)7.G.B.5Use facts about supplementary, complementary, vertical, and adjacent angles to solve for an unknown angle (Combining the adjacent base angles $\angle OBA$ and $\angle OBC$ that $OB$ splits $\angle ABC$ into, giving $\angle ABC=50^\circ$.)
⭐ Draw the radii from the center: every radius has the same length, so each triangle is isosceles, and their equal base angles let you build the angle you want by adding pieces.
⭐ Draw the radii from the center: every radius has the same length, so each triangle is isosceles, and their equal base angles let you build the angle you want by adding pieces.
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