Question

Answer 1

Kẻ đường cao AH

Xét tam giác AHB vuông tại H có \(\sin\widehat{B}=\sin40^0=\dfrac{AH}{AB}\approx0,64\Rightarrow AH\approx3,2\left(cm\right)\)

\(\cos\widehat{B}=\cos40^0=\dfrac{BH}{AB}\approx0,77\Rightarrow BH\approx3,85\left(cm\right)\)

Xét tam giác AHC vuông tại H có

\(\sin\widehat{C}=\sin65^0=\dfrac{AH}{AC}\approx0,91\Rightarrow AC=\dfrac{3,2}{0,91}\approx3,52\left(cm\right)\)

\(\cos\widehat{C}=\cos65^0=\dfrac{CH}{AC}\approx0,42\Rightarrow CH=0,42\cdot3,52\approx1,48\left(cm\right)\)

Vậy \(AC\approx3,52;BC=CH+BH\approx1,48+3,85=5,33\left(cm\right)\)