Răspuns:
[tex]\boldsymbol{ \red{ \hat{A} = 75^{\circ},\hat{B} = 45^{\circ},\hat{C} = 60^{\circ}}}[/tex]
[tex]\boldsymbol{ \red{AB = 27\sqrt{2} \ cm, BC = 9(3 + \sqrt{3}) \ cm}} \\[/tex]
Explicație pas cu pas:
AC = 18√3 cm
[tex]\sin C = \dfrac{\sqrt{3}}{2} \Rightarrow \hat{C} = 60^{\circ}[/tex]
[tex]\cos B = \dfrac{\sqrt{2}}{2} \Rightarrow \hat{B} = 45^{\circ}[/tex]
[tex]\hat A + \hat B + \hat C = 180^{\circ}[/tex]
[tex]\hat A = 180^{\circ} - (60^{\circ} + 45^{\circ}) = 180^{\circ} - 105^{\circ} = 75^{\circ}[/tex]
Construim înălțimea AD⊥BC, D∈BC
[tex]\Delta ADC: \sin \hat C = \dfrac{AD}{AC} \Rightarrow AD = \dfrac{\sqrt{3}}{2} \cdot 18\sqrt{3} = 27[/tex]
[tex]\cos \hat C = \dfrac{CD}{AC} \Rightarrow \cos 60^{\circ} = \dfrac{CD}{18\sqrt{3}} \Rightarrow CD = \dfrac{1}{2} \cdot 18\sqrt{3} = 9\sqrt{3} \ cm[/tex]
[tex]\Delta ADB: \sin \hat B = \dfrac{AD}{AB} \Rightarrow \sin 45^{\circ} = \dfrac{27}{AB} \Rightarrow \dfrac{1}{\sqrt{2}} = \dfrac{27}{AB} \Rightarrow \bf AB = 27\sqrt{2} \ cm[/tex]
ΔADB este dreptunghic isoscel ⇒ BD ≡ AD ⇒ BD = 27 cm
[tex]BC = BD + CD = 9\sqrt{3} + 27 = \bf 9(3 + \sqrt{3}) \ cm[/tex]
