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Math  /  Geometry

Question4 AMC 12 B Problems 8
Equilateral △ABC\triangle A B C with side length 14 is rotated about its center by angle θ\theta, where 0<θ≤60∘0<\theta \leq 60^{\circ}, to form △DEF\triangle D E F. See the figure. The area of hexagon ADBECFA D B E C F is 91391 \sqrt{3}. What is tan⁡θ\tan \theta ? 如图所示, 边长为 14 的等边三角形 △ABC\triangle A B C 绕其中心旋转 θ\theta 度得到 △DEF\triangle D E F, 其中 0<θ≤60∘0<\theta \leq 60^{\circ} 。若六边形 ADBECFA D B E C F 的面积为 91391 \sqrt{3}, 请问 tan⁡θ\tan \theta 的值是多少?

Studdy Solution
解方程求出 tan⁡θ\tan \theta：
化簡方程：
98−98θ60=91 98 - \frac{98\theta}{60} = 91
98−91=98θ60 98 - 91 = \frac{98\theta}{60}
7=98θ60 7 = \frac{98\theta}{60}
θ=7×6098 \theta = \frac{7 \times 60}{98}
θ=42098 \theta = \frac{420}{98}
θ=21049 \theta = \frac{210}{49}
θ=307 \theta = \frac{30}{7}
計算 tan⁡θ\tan \theta：
由於 θ\theta 是一個小角度，tan⁡θ≈θ\tan \theta \approx \theta：
tan⁡θ≈307 \tan \theta \approx \frac{30}{7}
解得：
307 \boxed{\frac{30}{7}}

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