Cho HCN ABCD kẻ AH vg góc vs BD(H thuộc BD) biết BH=6cm AH=4,8cm tính sd của góc ADB (làm tròn đến độ) và chu vi HCN
Tham khảo câu trả lời từ Trợ lý ảo Lazi Để tính được góc ADB, ta sử dụng định lý cosin trong tam giác AHB: \(cos(\angle AHB) = \frac{AH^2 + BH^2 - AB^2}{2 \cdot AH \cdot BH}\) \(cos(\angle AHB) = \frac{4,8^2 + 6^2 - AB^2}{2 \cdot 4,8 \cdot 6}\) \(cos(\angle AHB) = \frac{23,04 + 36 - AB^2}{28,8}\) \(cos(\angle AHB) = \frac{59,04 - AB^2}{28,8}\) Vì góc AHB là góc nhọn nên \(cos(\angle AHB) > 0\), suy ra: \(59,04 - AB^2 > 0\) \(AB^2 < 59,04\) \(AB < \sqrt{59,04}\) \(AB < 7,68\) Do đó, chu vi HCN là: \(HCN = AB + BC + CD + DA\) \(HCN = AB + 2 \cdot BH + 2 \cdot AH\) \(HCN = 7,68 + 2 \cdot 6 + 2 \cdot 4,8\) \(HCN = 7,68 + 12 + 9,6\) \(HCN = 29,28\) cm Vậy chu vi của HCN là 29,28 cm. 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