Đáp án:
$a)\sqrt{17}\\ b) \sqrt{33}$
Giải thích các bước giải:
$a) \Delta ABC, \tan \widehat{A}=2\sqrt{2} >0\\ \Rightarrow \widehat{A} \in (0^\circ;90^\circ)\\ \Rightarrow \cos \widehat{A} >0\\ \dfrac{1}{\cos^2 \widehat{A}}=\tan^2 \widehat{A}+1\\ \Rightarrow \cos^2 \widehat{A}=\dfrac{1}{\tan^2 \widehat{A}+1}\\ \Rightarrow \cos \widehat{A}= \sqrt{\dfrac{1}{\tan^2 \widehat{A}+1}}= \dfrac{1}{3}$
Áp dụng định lý $\cos:$
$BC^2=AB^2+AC^2-2AB.AC.\cos \widehat{A}\\ \Rightarrow BC=\sqrt{AB^2+AC^2-2AB.AC.\cos \widehat{A}}=\sqrt{17}$
$b) \Delta ABC, \tan \widehat{A}=-2\sqrt{2} <0\\ \Rightarrow \widehat{A} \in (90^\circ;180^\circ)\\ \Rightarrow \cos \widehat{A} <0\\ \dfrac{1}{\cos^2 \widehat{A}}=\tan^2 \widehat{A}+1\\ \Rightarrow \cos^2 \widehat{A}=\dfrac{1}{\tan^2 \widehat{A}+1}\\ \Rightarrow \cos \widehat{A}= -\sqrt{\dfrac{1}{\tan^2 \widehat{A}+1}}=- \dfrac{1}{3}$
Áp dụng định lý $\cos:$
$BC^2=AB^2+AC^2-2AB.AC.\cos \widehat{A}\\ \Rightarrow BC=\sqrt{AB^2+AC^2-2AB.AC.\cos \widehat{A}}=\sqrt{33}$
