Question

12,13,14,15 și 16 va rog

Answer 1

   
[tex]\displaystyle\\ 12)\\ \sin^2 30^o + \cos^2 30^o =\boxed{1}~~~\texttt{Formula fundamentala in trigonometrie.}\\\\ 13)\\ \sin 45^o + \cos45^o = \frac{ \sqrt{2}}{2} +\frac{ \sqrt{2}}{2} = \frac{ 2\sqrt{2}}{2} = \boxed{\sqrt{2}}\\\\ 14)\\ \sin60^o\cdot \cos30^o -\sin60^o\cdot \sin30^o =\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2}- \frac{\sqrt{3}}{2}\cdot \frac{1}{2} =\\\\ =\frac{3}{4}- \frac{\sqrt{3}}{4}= \boxed{\frac{3-\sqrt{3}}{4}} [/tex]


[tex]\displaystyle \\ 15)\\ _\texttt{Din A ducem perpendiculara AD pe BC cu D inclus in BC.}\\ \text{In triunghiul dreptunghic ABD avem:}\\ AB=12~cm={\bf ipotenuza}\\ BD=\frac{BC}{2}=\frac{12\sqrt{3}}{2}=6\sqrt{3}~cm={\bf cateta}\\\\ \Longrightarrow~\cos B=\frac{BD}{AB}=\frac{6\sqrt{3}}{12}=\frac{\sqrt{3}}{2} ~\Longrightarrow~\sphericalangle B=\boxed{30^o}\\\\ \sphericalangle C=\sphericalangle B=\boxed{30^o}\\\\ \sphericalangle A=180^o-\sphericalangle B-\sphericalangle C=180^o-30^o-30^o=\boxed{120^o} [/tex]


[tex]\displaystyle\\ 16a)\\ A\boxed{} = l^2 = 4^2 = \boxed{\bf 16~cm^2}\\ 16b)\\ AC = \sqrt{AB^2 + BC^2}= \sqrt{4^2 + 4^2}=\sqrt{16 + 16}=\sqrt{2\times16 }=\boxed{4 \sqrt{2} ~cm}[/tex]