Question

Dau de toate. repede ex : 12 , 13 !
Va rog ​

Answer 1

 

$\displaystyle\bf\\12)\\Se~d\u a:\\\Delta ABC~dreptunghic\\m(\sphericalangle A)=90^o\\AB=12~cm\\tg~B=1\\\\Se~cere:\\AC=?\\BC=?\\cosB=?\\sinC=?\\ctg B\\tg C\\\\Rezolvare:\\tgB=1~~\implies~<\!\!B=45^o\\<\!\!C=90-45=45\\\implies~\Delta ABC~este~dreptunghic~isoscel\\\implies~AC=AB=12~cm\\BC=\sqrt{AB^2+AC^2}=\sqrt{12^2+12^2}=\sqrt{2\times12^2}=12\sqrt{2}\\\\cosB=cos45^o=\frac{\sqrt{2}}{2}\\\\sinC=sin45^o=\frac{\sqrt{2}}{2}\\\\ctgB=ctg45^o=1\\\\tgC=tg45^o=1$

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$\displaystyle\bf\\13)\\Se~da:\\\Delta ABC~~dreptunghic, m(\sphericalangle A)=90^o\\Cazurile:~a;~b;~c;~d\\\\Se~cere:\\Perimetrul~pentru~fiecare~caz.\\\\\\a)\\AB=16~cm~~si~~sinC=\frac{8}{17}\\\\sinC=\frac{AB}{BC}=\frac{8}{17}\\\\\frac{16}{BC}=\frac{8}{17}\\\\BC=\frac{16\times17}{8}=2\times17=34~cm\\\\AC=\sqrt{BC^2-AB^2}=\\\\=\sqrt{34^2-16^2}=\sqrt{1156-256}=\sqrt{900}=30~cm\\\\\boxed{\bf P=AB+AC+BC=16+30+34=80~cm}$

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$\displaystyle\bf\\b)\\AC=20~~si~~tgB=\frac{5}{12}\\\\tgB=\frac{AC}{AB}=\frac{5}{12}\\\\\frac{20}{AB}=\frac{5}{12}\\\\AB=\frac{20\times12}{5}=4\times12=48~cm\\\\BC=\sqrt{AB^2+AC^2}=\\\\=\sqrt{48^2+20^2}=\sqrt{2304+400}=\sqrt{2704}=52~cm\\\\\boxed{\bf P=AB+AC+BC=48+20+52=120~cm}$

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$\displaystyle\bf\\c)\\BC=15~cm~~si~~cosC=\frac{3}{5}\\\\cosC=\frac{AB}{BC}=\frac{3}{5}\\\\\frac{AB}{15}=\frac{3}{5}\\\\AB=\frac{15\times3}{5}=3\times3=9~cm\\\\AC=\sqrt{BC^2-AB^2}=\sqrt{15^2-9^2}=\sqrt{225-81}=\sqrt{144}=12~cm\\\\\boxed{\bf P=AB+AC+BC=9+12+15=36~cm}$

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$\displaystyle\bf\\d)\\AB=18~cm~~si~~ctgB=\frac{3}{4}\\\\ctgB=\frac{AB}{AC}=\frac{3}{4}\\\\\frac{18}{AC}=\frac{3}{4}\\\\AC=\frac{18\times4}{3}=6\times4=24~cm\\\\BC=\sqrt{AB^2+AC^2}=\sqrt{18^2+24^2}=\sqrt{324+576}=\sqrt{900}=30~cm\\\\\boxed{\bf P=AB+AC+BC=18+24+30=72~cm}$