AMC 8 · 1999 · #21
Grade 7 geometry-2d
Pick an answer.
AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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° and 40° marks, and a second one that holds the 110° 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.
First subproblem: the triangle holding 100° and 40°. Its angles sum to 180°, so the third angle is 40°.
Once two angles of a triangle are known, the third is just 180° minus their sum.
7.G.B.5Identify SubproblemsThat 40° sits on the side shared with the next triangle, so it is also an interior angle of the triangle holding angle A and 110°.
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.
4.G.A.1Draw A DiagramClose the second triangle with the angle-sum rule: A, 110°, and the transferred 40° give A = 30°.
The same rule that closed the first triangle closes the second — two passes of "angles add to 180°" is the whole proof.
7.G.B.5Identify SubproblemsTwo small triangles, one shared side. The first triangle pins down a 40° angle (180 - 100 - 40); reusing it in the second triangle forces A = 180 - 110 - 40 = 30° — choice (B).