Cho tứ giác ABCD có AB = AC = AD = 20cm, $\hat{B}={60}^0$ và $\hat{A}={90}^0$. Kẻ BE DC kéo dài.

Tính BE?

A. sin B = $\frac{AH}{AB}$

B. cos C = $\frac{AC}{BC}$

C. tan B = $\frac{AC}{AB}$

D. tan C = $\frac{AH}{AC}$

A. $A{H}^2=BH.CH$

B. $A{B}^2=BH.BC$

C. $\frac{1}{A{H}^2}=\frac{1}{A{B}^2}+\frac{1}{A{C}^2}$

D. AH. AB = BC. AC

A. BH = 2cm, CH = 3,2cm, AC = 4cm, AH = 2,4cm

B. BH = 1,8cm; CH = 3,2cm; AC = 4cm; AH = 2,4cm

C. BH = 1,8cm; CH = 3,2cm; AC = 3cm; AH = 2,4cm

D. BH = 1,8cm; CH = 3,2cm; AC = 4cm; AH = 4,2cm

A. $\frac{\sqrt{93}}{14}$

B. $14\sqrt{93}$

C. $\frac{14\sqrt{93}}{93}$

D. $\frac{14}{17}$

A. $\hat{B}={45}^0$; $\hat{C}={45}^0$; AH = $\sqrt{3}$

B. $\hat{B}={50}^0$; $\hat{C}={40}^0$; AH = 2

C. $\hat{B}={30}^0$; $\hat{C}={60}^0$; AH = 4

D. $\hat{B}={60}^0$; $\hat{C}={30}^0$; AH = 2

A. AC = 37,2cm; AB = 33,4cm; $\hat{C}={32}^0$

B. AC = 37,2cm; AB = 33,5cm; $\hat{C}={42}^0$

C. AB = 37,2cm; AC = 33,5cm; $\hat{C}={42}^0$

D. AC = 37,2cm; AB = 33,5cm; $\hat{C}={42}^0$

A. $\frac{1}{3}$

B. 3

C. $\sqrt{3}$

D. $\frac{1}{2}$