AMC 8 · 1999 · #21
Grade 7 geometry-2dProblem
The degree measure of angle A is
Pick an answer.
AMC 8 1999 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: A self-intersecting closed figure (a five-pointed star-like polygon) shows three labeled angles inside its small triangular regions: $40^\circ$ at the bottom-left tip, $100^\circ$ in a middle triangle, and $110^\circ$ in a triangle near vertex $A$. Find the degree measure of angle $A$.
Givens: Three marked angles in the figure: $40^\circ$, $100^\circ$, and $110^\circ$; Angle $A$ is one vertex of the outer polygon; Answer choices: (A) $20$, (B) $30$, (C) $35$, (D) $40$, (E) $45$
Unknowns: The degree measure of angle $A$
Understand
Restated: A self-intersecting closed figure (a five-pointed star-like polygon) shows three labeled angles inside its small triangular regions: $40^\circ$ at the bottom-left tip, $100^\circ$ in a middle triangle, and $110^\circ$ in a triangle near vertex $A$. Find the degree measure of angle $A$.
Givens: Three marked angles in the figure: $40^\circ$, $100^\circ$, and $110^\circ$; Angle $A$ is one vertex of the outer polygon; Answer choices: (A) $20$, (B) $30$, (C) $35$, (D) $40$, (E) $45$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram
The figure looks tangled, but the three marked angles each sit inside a small triangle formed by the crossing segments. Tool #7 (Identify Subproblems) breaks the picture into just two of those triangles: one that holds the $100^\circ$ and $40^\circ$ marks, and a second one that holds the $110^\circ$ mark together with angle $A$. The two triangles share a side, so the third angle of the first triangle reappears as a known angle in the second. Tool #1 (Draw a Diagram) is the bookkeeping partner — mark each angle on the figure as you find it so the transfer between the two triangles is visible. No algebra needed; the triangle-angle-sum rule does all the work.
Execute — Answer: B
7.G.B.5 Step 1 - Pick out the first subproblem: the small triangle that contains the $100^\circ$ angle and the $40^\circ$ angle.
- Its three interior angles must add to $180^\circ$, so the unmarked third angle is forced.
💡 Once two angles of a triangle are known, the third is just $180^\circ$ minus their sum.
4.G.A.1 Step 2 - Mark that newly found $40^\circ$ on the diagram at the shared vertex.
- The same segment that forms one side of the first triangle also forms a side of the second (the triangle that contains angle $A$ and the $110^\circ$ mark), so the $40^\circ$ angle is one of the interior angles of that second triangle too.
💡 The two triangles share a side, so the angle that sits along that shared side is the same in both triangles. Drawing it on the figure makes the reuse obvious.
7.G.B.5 Step 3 Solve the second subproblem: apply the triangle-angle-sum rule to the triangle holding angle $A$, the $110^\circ$ mark, and the transferred $40^\circ$.
💡 The same rule that closed the first triangle closes the second — two passes of "angles add to $180^\circ$" is the whole proof.
7.G.B.5 Pick out the first subproblem: the small triangle that contains the $100^\circ$ 4.G.A.1 Mark that newly found $40^\circ$ on the diagram at the shared vertex. The same s 7.G.B.5 Solve the second subproblem: apply the triangle-angle-sum rule to the triangle h Review
Reasonableness: Both triangles close cleanly: $100 + 40 + 40 = 180$ for the first triangle, and $110 + 40 + 30 = 180$ for the second. The answer $30^\circ$ is also positive and well under $180^\circ$, as any triangle interior angle must be — and it matches choice (B). A quick second check: $A$ shares its triangle with a wide $110^\circ$ angle, so the remaining two angles must add to only $70^\circ$; splitting that as $40^\circ + 30^\circ$ fits, and any answer larger than $70^\circ$ would have been impossible.
Alternative: Tool #5 (Look for a Pattern) via the exterior-angle theorem. The $110^\circ$ mark is the exterior angle of the first triangle at the shared vertex (because the segment continues straight through). The exterior-angle rule says it equals the sum of the two remote interior angles, so $110^\circ = 100^\circ + (\text{angle next to } A)$ would not match — instead use it on the second triangle: the third angle of the first triangle ($40^\circ$) plus $A$ equals the exterior angle at the far vertex, which the figure shows is supplementary to $110^\circ$, giving $40^\circ + A = 70^\circ$ and $A = 30^\circ$. Either bookkeeping route lands on (B).
CCSS standards used (min grade 7)
7.G.B.5Use facts about supplementary, complementary, vertical, and adjacent angles in a multi-step problem to write and solve simple equations for an unknown angle in a figure (Applying the triangle-angle-sum rule twice — first to find the third angle of the $(100^\circ, 40^\circ, ?)$ triangle, then to find $A$ inside the $(A, 110^\circ, 40^\circ)$ triangle.)4.G.A.1Draw points, lines, line segments, rays, angles, and identify these in two-dimensional figures (Recognizing the shared side between the two small triangles so the $40^\circ$ third angle of the first triangle reappears as an interior angle of the second triangle.)
⭐ Two small triangles, one shared side. The first triangle pins down a $40^\circ$ angle ($180 - 100 - 40$); reusing it in the second triangle forces $A = 180 - 110 - 40 = 30^\circ$ — choice (B).
⭐ Two small triangles, one shared side. The first triangle pins down a $40^\circ$ angle ($180 - 100 - 40$); reusing it in the second triangle forces $A = 180 - 110 - 40 = 30^\circ$ — choice (B).
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