Question

\(\Delta ABC\), \(\widehat{A}< 90^0\), AB = 16 cm, AC = 14cm, \(\widehat{B}=60^0\). Tính BC

Answer 1

Xét ΔABC có \(\cos B=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}=\dfrac{16^2-14^2+BC^2}{2\cdot16\cdot BC}\)

\(\Leftrightarrow\dfrac{BC^2+60}{32BC}=\dfrac{1}{2}\)

\(\Leftrightarrow2BC^2-32BC+60=0\)

\(\Leftrightarrow BC^2-16BC+30=0\)

\(\Leftrightarrow BC^2-16BC+64-34=0\)

\(\Leftrightarrow\left[{}\begin{matrix}BC-8=\sqrt{34}\\BC-8=-\sqrt{34}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}BC=\sqrt{34}+8\left(cm\right)\\BC=-\sqrt{34}+8\left(cm\right)\end{matrix}\right.\)