English Quadrilateral ABCD satisfies ∠ ABD = ∠ ADB = 45°, ∠ ACB = 30°, ∠ ABC = 127.5°. Find ∠ ACD°. Let XY represent the line segment XY. For convenience, we consider AB=1. Triangle ABD is a right isosceles triangle with ∠A=90°, so according to the Pythagorean theorem, BD=√2. Regarding triangle ABC, by the law of sines, AB/sin30° = AC/sin127.5° = BC/sin22.5°, thus BC = (sin22.5°/sin30°) * AB = 2sin22.5°. Therefore, BD/BC = √2/(2sin22.5°) = 2cos22.5° = 2sin67.5°...[1] Let ∠BDC=x, then according to the law of sines in triangle BDC, BD/BC = sin∠BCD/sin∠BDC = sin(97.5°-x)/sin x. From [1], BD/BC = 2sin67.5°, thus x is 30°. Therefore, ∠BDC=30° and using simple angle calculations, ∠ACD=37.5°... (Google翻訳)
記述があまりに稚拙 （いろいろ計算を省略しています。） 答えがあっている自信がない 正弦定理で幾何の問題を解いたことが一度もないため。
