🇺🇸 AMC 8 ⇄ switch contest
1997 AJHSME

Problem 12

Problem 12 · 1997 AJHSME Medium
Geometry & Measurement angle-chase
Figure for AJHSME 1997 Problem 12
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Answer: D — 35°.
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Hint 1 of 2
∠4 lives in the right triangle, but you can't touch it yet. Start where ALL the numbers are — the left triangle — and let each angle hand you the next one, like dominoes.
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Hint 2 of 2
Angle-chasing toolkit: angles in a triangle sum to 180°, and two angles on a straight line are supplementary (sum to 180°). Chain them.
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Approach: chase angles step by step into the target triangle
  1. Left triangle has 40° (apex) and 70° (base), so the third angle ∠1 = 180° − 40° − 70° = 70°.
  2. ∠1 and ∠2 sit on the straight base line, so ∠2 = 180° − 70° = 110°.
  3. Now the right triangle's three angles are ∠2, ∠3, ∠4: ∠3 + ∠4 = 180° − 110° = 70°. Since ∠3 = ∠4 (given), each is 70° ÷ 2 = 35°.
  4. Why this transfers: angle problems are dominoes — you can't reach the far angle directly, so start from the fully-determined triangle and propagate one relation at a time.
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