Question

5 si 6
Dau coroana si 28 de puncte.
Materie clasa a 10-a.
Multumesc anticipat.​

Answer 1

 

$\displaystyle\bf\\5) (Vezi~desenul~atasat.)\\\\\overrightarrow{\bf AB}+\overrightarrow{\bf AC}+\overrightarrow{\bf AD}=?\\\\\overrightarrow{\bf AB}+\overrightarrow{\bf AD}=\overrightarrow{\bf AC}~~~~~(Conform~metodei~paralelogramului)\\\\\textbf{Patratul este un caz particular de paralelogram.}\\\\Lungimea~vectorului~\overrightarrow{\bf AC}=lungimea~diagonalei~patratului.\\\\L=latura~patratului\\\\luungimea~lui~~\overrightarrow{\bf AC}=L\sqrt{2}$

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$\displaystyle\bf\\\overrightarrow{\bf AB}+\overrightarrow{\bf AC}+\overrightarrow{\bf AD}= \underbrace{\overrightarrow{\bf AB}+\overrightarrow{\bf AD}}_{=\overrightarrow{\bf AC}}+\overrightarrow{\bf AC}=\overrightarrow{\bf AC}+\overrightarrow{\bf AC}=2\overrightarrow{\bf AC}\\\\\\Lungimea~vectorului~2\overrightarrow{\bf AC}=2\times L\sqrt{2}=\boxed{\bf2L\sqrt{2}}$

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6)

$\displaystyle\bf\\sin\,105^o=sin(60^o+45^o)\\\\Folosim~formula\!:\\\\sin(\propto+\beta)=sin\propto cos\beta+sin\beta cos\propto\\\\sin\,105^o=sin(60^o+45^o)=sin60^ocos45^o+sin45^ocos60^o=\\\\=\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\\\\=\frac{\sqrt{3}\times\sqrt{2}}{2\times2}+\frac{\sqrt{2}\times1}{2\times2}=\\\\=\frac{\sqrt{3\times2}}{4}+\frac{\sqrt{2}}{4}=\\\\=\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}=\boxed{\bf\frac{\sqrt{6}+\sqrt{2}}{4}}$