Giải giúp mik vs nhe :))

Đáp án + Giải thích các bước giải:

$\textbf{a}\bigg)$

$\widehat{B} = 90^\circ - \widehat{B} = 90^\circ - 30^\circ = 60^\circ$

Xét $\triangle ABC$ vuông tại $A$, ta có:

$\sin \widehat{ACB} = \dfrac{AB}{BC}$

$\Rightarrow \sin 30^\circ = \dfrac{9}{BC}$

$\Rightarrow BC = \dfrac{9}{\sin 30^\circ} = 18$

Mặt khác, ta có:

$\cos \widehat{ACB} = \dfrac{AC}{BC}$

$\Rightarrow \cos 30^\circ = \dfrac{AC}{18}$

$\Rightarrow AC = 18 \cos 30^\circ = 9\sqrt{3}(cm)$

$\textbf{b}\bigg)$

Xét $\triangle AHC$ vuông tại $H$, ta có:

$\sin \widehat{ACH} = \dfrac{AH}{AC}$

$\Rightarrow \sin 30^\circ = \dfrac{AH}{9\sqrt{3}}$

$\Rightarrow AH = 9\sqrt{3} \sin 30^\circ = \dfrac{9}{2}\sqrt{3}(cm)$

Mà $AH^2 + HC^2 = AC^2($định lý Pythagore$)$

$\Rightarrow \bigg(\dfrac{9}{2}\sqrt{3}\bigg)^2 + HC^2 = (9\sqrt{3})^2$

$\Rightarrow HC^2 = 243 - \dfrac{243}{4} = \dfrac{729}{4}$

$\Rightarrow HC = \dfrac{27}{2}(cm)$

$\textbf{c}\bigg)$

Xét $\triangle ABC$, có phân giác $AD$

$\Rightarrow \dfrac{DB}{DC} = \dfrac{AB}{AC}$

$\Rightarrow \dfrac{18 - DC}{DC} = \dfrac{9}{9\sqrt{3}}$

$\Rightarrow 9\sqrt{3}(18 - DC)= 9DC$

$\Rightarrow 162\sqrt{3}DC - 9\sqrt{3}DC = 9DC$

$\Rightarrow (9 + 9\sqrt{3})DC = 162\sqrt{3}$

$\Rightarrow DC = \dfrac{162\sqrt{3}}{9 + 9\sqrt{3}} = 27 - 9\sqrt{3}(cm)$

$\Rightarrow HD = HC - DC = \dfrac{27}{2} - (27 - 9\sqrt{3}) = \dfrac{18\sqrt{3} - 27}{2}(cm)$

Xét $\triangle HDA$ vuông tại $H$, ta có:

$AH^2 + HD^2 = AD^2($định lý Pythagore$)$

$\Rightarrow AD^2 = \bigg(\dfrac{9}{2}\sqrt{3}\bigg)^2 + \bigg(\dfrac{18\sqrt{3} - 27}{2}\bigg)^2$

$\Rightarrow AD^2 = \dfrac{243}{4} + \dfrac{1701 - 972\sqrt{3}}{4}$

$\Rightarrow AD^2 = \dfrac{1944 - 972\sqrt{3}}{4} = 486 - 243\sqrt{3} = 121,5(4 - 2\sqrt{3})$

$\Rightarrow AD = \sqrt{121,5} \cdot \sqrt{4 - 2\sqrt{3}} = \dfrac{9\sqrt{6}}{2}(\sqrt{3} - 1)$

$\Rightarrow AD = \dfrac{27\sqrt{2} - 9\sqrt{6}}{2} \approx 8,07(cm)$