Question

Answer 1

\(a,\) Ta có \(\cos\widehat{B}=\cos35^0=\dfrac{AB}{BC}\approx0,8\Leftrightarrow BC\approx\dfrac{5}{0,8}=6,25\left(cm\right)\)

Áp dụng HTL: \(\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BC}\\AH^2=BH\cdot HC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}BH=4\left(cm\right)\\AH=\sqrt{4\left(6,25-4\right)}=3\left(cm\right)\end{matrix}\right.\)

\(b,\) Áp dụng Pytago \(AC=\sqrt{BC^2-AB^2}=3,75\left(cm\right)\)

Ta có \(\widehat{B}+\widehat{C}=90^0\Rightarrow\widehat{C}=90^0-35^0=55^0\)