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Cho hình chữ nhật ABCD

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
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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.

Để tính góc ADB, ta có:

\(180^\circ = \angle AHB + \angle BHD + \angle DHC + \angle CHA\)

\(180^\circ = \angle AHB + \angle BHD + \angle DHC + \angle CHA\)

\(180^\circ = \angle AHB + \angle BHD + \angle DHC + (180^\circ - \angle AHB)\)

\(180^\circ = \angle BHD + \angle DHC\)

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