Question

In triunghiul abc cu m (abc)=135 de grade ab =18radical 2 iar bc=ab radical 2 si ad perpendicular bc d apartine bc. Calculati lungimea seg ad lungimea latirii bc si tg c

Answer 1

   
[tex]\displaystyle\\ \text{Se da:}\\ \Delta ABC \text{ cu } m(\sphericalangle ABC) = 135^o\\ AB = 18\sqrt{2}~cm\\ BC=AB\sqrt{2}~cm\\ AD \perp BC,~D\in BC\\\\ \text{Se cere:}\\ AD=?\\ AC=? ~~~\text{(nu BC)}\\ \text{tg}\,C=? [/tex]


[tex]\displaystyle\\ \text{Rezolvare:}\\\\ BC = AB\sqrt2 \cdot \sqrt2 = 18\sqrt2 \cdot \sqrt2 = 18 \cdot 2 = 36~cm\\ AC^2 = AB^2 + BC^2 -2AB \cdot BC \cdot \cos(ABC)=\\ = (18\sqrt{2})^2 + 36^2 - 2\cdot 18\sqrt{2} \cdot 36 \cdot \cos 135^o=\\ = (18\sqrt{2})^2 + 36^2 - 2\cdot 18\sqrt{2} \cdot 36 \cdot \cos (180^o-135^o)=\\ = (18\sqrt{2})^2 + 36^2 - 2\cdot 18\sqrt{2} \cdot 36 \cdot (- \cos (45)=\\\\ = (18\sqrt{2})^2 + 36^2 + 2\cdot 18\sqrt{2} \cdot 36 \cdot \frac{\sqrt{2}}{2} = [/tex]

[tex]\displaystyle\\ = 324 \cdot 2 + 1296 + 18\sqrt{2} \cdot 36 \cdot \sqrt{2} =\\\\ = 648 + 1296 + 18\cdot 2\cdot 36 = \\ = 648 + 1296 + 36\cdot 36 =\\ = 648 + 1296 + 1296 = 3240 \\\\ AC^2=3240\\ AC = \sqrt{3240} =\sqrt{324\cdot 10} =\boxed{\bf18\sqrt{10}} [/tex]


[tex]\displaystyle\\ \text{In }~\Delta ABD ~\text{ avem:}\\ \sphericalangle ABD = 180 - \sphericalangle ABC=180 - 135 = 45^o \\ \Longrightarrow~~\Delta ABD \text{ este triunghi dreptunghic isoscel.}\\\\ AD=BD=AB\sin45^o =18 \sqrt{2} \cdot \frac{\sqrt{2}}{2} = \frac{18\cdot 2}{2} = \boxed{\bf 18~cm} \\\\ \text{In }~\Delta ADC ~\text{ avem:}\\ AD=18~cm = cateta\\ CD = DB + BC = 18 + 36 = 54 cm=cateta \\\\ \text{tg }C = \frac{AD}{CD} = \frac{18}{54} = \boxed{\bf \frac{1}{3}}[/tex]