a)
\[
\sin C = \frac{AB}{BC} \Rightarrow \sin 30^\circ = \frac{AB}{18}
\]
\[
\frac{1}{2} = \frac{AB}{18} \Rightarrow AB = 18 \cdot \frac{1}{2} = 9 \text{ cm}
\]
\[
\cos C = \frac{CA}{BC} \Rightarrow \cos 30^\circ = \frac{CA}{18}
\]
\[
\frac{\sqrt{3}}{2} = \frac{CA}{18} \Rightarrow CA = 18 \cdot \frac{\sqrt{3}}{2} = 9\sqrt{3} \text{ cm}
\]
\[
\angle A + \angle B + \angle C = 180^\circ \Rightarrow 90^\circ + \angle B + 30^\circ = 180^\circ
\]
\[
\angle B = 60^\circ
\]
- \( AB = 9 \text{ cm} \)
- \( CA = 9\sqrt{3} \text{ cm} \)
- \( \angle B = 60^\circ \)
---
b)
\[
HC = AB \cdot \tan C = 9 \cdot \tan 30^\circ = 9 \cdot \frac{1}{\sqrt{3}}
\]
\[
HC = \frac{9}{\sqrt{3}} = 3\sqrt{3} \text{ cm}
\]
\[
\cos C \cdot \cos B = \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \frac{\sqrt{3}}{4}
\]
\[
\frac{HC}{BC} = \frac{3\sqrt{3}}{18} = \frac{\sqrt{3}}{6}
\]
So sánh
\[
HC = AH \cdot \tan C \text{ và }\frac{HC}{BC} = \frac{HC}{18}
\]
c)
\[
\angle ABK = \angle KBC \quad \text{và} \quad \angle ABE = \angle EBC
\]
\[
\angle ABK + \angle ABE = 60^\circ \Rightarrow KE \parallel BC
\]
d)
\[
AH = AB \cdot \sin C = 9 \cdot \frac{1}{2} = 4.5 \text{ cm}
\]
\[
S_{AKB} = \frac{1}{2} \cdot AB \cdot AH = \frac{1}{2} \cdot 9 \cdot 4.5 = 20.25 \text{ cm}^2
\]
\[
S_{AKBE} = S_{AKB} + S_{ABE} = 20.25 \cdot 2 = 40.5 \text{ cm}^2
\]