Câu hỏi của koquantam

Question

Giải tam giác ABC vuông tại A , biết AB = 30 cm và $\widehat{C}$= 30 độ

Duc Dinh · Accepted Answer

$\widehat{C}=30^0\Rightarrow\widehat{B}=60^0$

$sinC=\frac{AB}{BC}\Leftrightarrow BC=\frac{AB}{sinC}=\frac{10}{sin30^0}=20\left(cm\right)$

$\Rightarrow AC=\sqrt{BC^2-AB^2}=\sqrt{20^2-10^2}=10\sqrt{3}\left(cm\right)$ (PITAGO)

Câu trả lời:

$\widehat{C}=30^0\Rightarrow\widehat{B}=60^0$

$sinC=\frac{AB}{BC}\Leftrightarrow BC=\frac{AB}{sinC}=\frac{10}{sin30^0}=20\left(cm\right)$

$\Rightarrow AC=\sqrt{BC^2-AB^2}=\sqrt{20^2-10^2}=10\sqrt{3}\left(cm\right)$ (PITAGO)

Đông Phương Lạc · Answer

$\widehat{C}=30^0\Rightarrow\widehat{B}=60^0$

$\sin C=\frac{AB}{BC}\Leftrightarrow BC=\frac{AB}{\sin C}=\frac{10}{\sin30^0}=20\left(cm\right)$

$\Rightarrow AC=\sqrt{BC^2-AB^2}=\sqrt{20^2-10^2}=10\sqrt{3}\left(cm\right)$

# Hok_tốt nha

Câu trả lời:

$\widehat{C}=30^0\Rightarrow\widehat{B}=60^0$

$\sin C=\frac{AB}{BC}\Leftrightarrow BC=\frac{AB}{\sin C}=\frac{10}{\sin30^0}=20\left(cm\right)$

$\Rightarrow AC=\sqrt{BC^2-AB^2}=\sqrt{20^2-10^2}=10\sqrt{3}\left(cm\right)$

# Hok_tốt nha