Để giải tam giác ABC vuông tại A với góc C = 30 độ và AB = 9 cm, ta thực hiện các bước sau:

### a) Giải tam giác ABC

1. **Tính góc B:**
- Tam giác ABC vuông tại A, nên góc A = 90 độ.
- Góc C = 30 độ.
- Suy ra góc B = 180 độ - góc A - góc C = 180 độ - 90 độ - 30 độ = 60 độ.

2. **Tính các cạnh còn lại:**
- AB là cạnh huyền, nên AB = 9 cm.
- Sử dụng định lý sin trong tam giác vuông:
- \( \sin C = \frac{BC}{AB} \)
- \( \sin 30^\circ = \frac{BC}{9} \)
- \( \frac{1}{2} = \frac{BC}{9} \)
- \( BC = 9 \times \frac{1}{2} = 4.5 \) cm.

- Sử dụng định lý cosin trong tam giác vuông:
- \( \cos C = \frac{AC}{AB} \)
- \( \cos 30^\circ = \frac{AC}{9} \)
- \( \frac{\sqrt{3}}{2} = \frac{AC}{9} \)
- \( AC = 9 \times \frac{\sqrt{3}}{2} = 4.5\sqrt{3} \) cm.

### b) Tính đường cao AH và đoạn BH

1. **Tính đường cao AH:**
- Trong tam giác vuông, đường cao từ đỉnh góc vuông đến cạnh huyền chia tam giác thành hai tam giác vuông nhỏ hơn đồng dạng với tam giác ban đầu.
- Sử dụng công thức đường cao trong tam giác vuông:
- \( AH = \frac{AC \times BC}{AB} \)
- \( AH = \frac{4.5\sqrt{3} \times 4.5}{9} \)
- \( AH = \frac{20.25\sqrt{3}}{9} \)
- \( AH = 2.25\sqrt{3} \) cm.

2. **Tính đoạn BH:**
- Sử dụng định lý Pythagore trong tam giác vuông ABH:
- \( BH = \sqrt{AB^2 - AH^2} \)
- \( BH = \sqrt{9^2 - (2.25\sqrt{3})^2} \)
- \( BH = \sqrt{81 - 15.1875} \)
- \( BH = \sqrt{65.8125} \)
- \( BH \approx 8.11 \) cm.

### c) Tính độ dài phân giác AD của tam giác ABC

1. **Sử dụng công thức phân giác trong tam giác vuông:**
- \( AD = \frac{2 \times AC \times BC}{AC + BC} \times \cos \left(\frac{\angle C}{2}\right) \)
- \( AD = \frac{2 \times 4.5\sqrt{3} \times 4.5}{4.5\sqrt{3} + 4.5} \times \cos \left(\frac{30^\circ}{2}\right) \)
- \( AD = \frac{2 \times 4.5\sqrt{3} \times 4.5}{4.5(\sqrt{3} + 1)} \times \cos 15^\circ \)
- \( AD = \frac{2 \times 4.5 \times 4.5\sqrt{3}}{4.5(\sqrt{3} + 1)} \times \cos 15^\circ \)
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