Đáp án:
Tham khảo
Giải thích các bước giải:
$\text{ Kẻ đường cao AH,CF,BD}$
Ta có:$sinA=\dfrac{BD}{AB}=\dfrac{BD}{a}=\dfrac{CF}{AC}=\dfrac{CF}{b}$
$sinB=\dfrac{CF}{AB}=\dfrac{CF}{a}=\dfrac{AH}{AB}=\dfrac{AH}{c}$
$sinC=\dfrac{AH}{AC}=\dfrac{AH}{b}=\dfrac{BD}{BC}=\dfrac{BD}{a}$
$⇒\dfrac{sinA}{sinB}=\dfrac{CF}{a}:\dfrac{CF}{b}⇒\dfrac{b}{sinB}=\dfrac{a}{SinA}. (1)$
$⇒\dfrac{sinC}{sinA}=\dfrac{BD}{a}:\dfrac{BD}{b}=\dfrac{c}{a}⇒\dfrac{SinC}{c}=\dfrac{a}{sinA}. (2)$
$\text{Từ (1) và (2)}$⇒$\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}$
