1972 AMC 12 Problem 14

Attempt Problem 14 of the 1972 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1972 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

14.

A triangle has angles of 30∘30^\circ and 45∘.45^\circ. If the side opposite the 45∘45^\circ angle has length 8,8, then the side opposite the 30∘30^\circ angle has length:
44
424\sqrt2
434\sqrt3
464\sqrt6
66
Answer: B
Concepts:law of sinestrigonometry
Small hint:
Relate the two sides with the Law of Sines
Big hint:
Use sin⁡30∘=12\sin30^\circ=\frac{1}{2} and sin⁡45∘=22\sin45^\circ=\frac{\sqrt2}{2}
Solution:
If ss is the side opposite 30∘,30^\circ, the Law of Sines gives ssin⁡30∘=8sin⁡45∘. \frac{s}{\sin30^\circ}=\frac8{\sin45^\circ}. Therefore s=81222=42. s=8\frac{\frac{1}{2}}{\frac{\sqrt2}{2}}=4\sqrt2.
Therefore, the correct answer is B.
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