$a) AC=16cm ; \hat{C}=42^o$
$∆ABC$ vuông tại $\hat{A}$
$→\hat{B}+\hat{C}=90^o$
$→ \hat{B}=90^o-\hat{C}=90^o-42^o=48^o$
$AB=AC.tanC=16.tan42^o≈36,66cm$
Áp dụng định lí Pytago trong tâm giác vuông $ABC$
Có $BC^2=AB^2+AC^2$
$→BC=\sqrt{AB^2+AC^2}=\sqrt{36,66^2+16^2}≈40cm$
$b) BC=25cm ; AC=15cm$
$∆ABC$ vuông tại $\hat{A}$ có
$sin{B}=\dfrac{AC}{BC}=\dfrac{15}{25}=0,6→\hat{B}≈36,9^o$
$\hat{B}+\hat{C}=90^o$
$→\hat{C}=90^o-36,9^o=53,1^o$
$AB=\sqrt{BC^2-AC^2}=\sqrt{25^2-15^2}=20cm$
$c) AC=12cm ; AB=14cm$
$∆ABC$ vuông tại $\hat{A}$ có
$tan{C}=\dfrac{AB}{AC}=\dfrac{14}{12}$
$→\hat{C}=49,4^o$
$\hat{B}+\hat{C}=90^o$
$→\hat{B}=90^o-49,4^o=40,6^o$
$BC=\sqrt{AB^2+AC^2}=\sqrt{14^2+12^2}≈18,4cm$
$d) BC=25cm; \hat{B}=48^o$
$∆ABC$ vuông tại $B$ có
$\hat{B}+\hat{C}=90^o$
$→\hat{C}=90^o-\hat{C}=90^o-48^o=42^o$
$AB=BC.sinC=25.sin42^o≈16,73cm$
$AC=BC.sinB=25.sin48^o≈18,59cm$
