Question

va rog 6 urgent rezolvarea ​

Answer 1

$m(\angle A)=90^\circ\Rightarrow\triangle ABC\text{ deptunghic \^in }\angle A\Rightarrow BC^2=AB^2+AC^2\\\Rightarrow (2AB)^2=AB^2+AC^2\Rightarrow AC^2=3AB^2\Rightarrow AC=\sqrt3AB$

Aplicam teorema sinusului:

$\displaystyle\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\Rightarrow\frac{a}{\sin A}=\frac{c}{\sin C}\\\Rightarrow \frac{BC}{\sin90^\circ}=\frac{AB}{\sin C}\Rightarrow \sin C=\frac{AB}{BC}=\frac{AB}{2AB}\Rightarrow \sin C=\frac12$

Aplicam teorema cosinusului:

$\cos B=\displaystyle\frac{a^2+c^2-b^2}{2\cdot a\cdot c}=\frac{BC^2+AB^2-AC^2}{2\cdot BC\cdot AB}=\\\\=\frac{4AB^2+AB^2-3AB^2}{2\cdot2AB\cdot AB}=\frac{2AB^2}{4AB^2}\Rightarrow \cos B=\frac12$

$\cos C=\displaystyle\frac{a^2+b^2-c^2}{2ab}=\frac{BC^2+AC^2-AB^2}{2\cdot BC\cdot AC}=\\=\frac{4AB^2+3AB^2-AB^2}{2\cdot 2AB\cdot\sqrt3AB}=\frac{6AB^2}{4\sqrt3AB^2}=\frac{3}{2\sqrt3}=\frac{3\sqrt3}{6}=\frac{\sqrt3} 2$

$tg B=\displaystyle\frac{AC}{AB}=\frac{\sqrt3AB}{AB}=\sqrt3$

$\displaystyle\frac{tgB+2\sin C}{\cos B+\cos C}=\frac{\sqrt3+2\cdot\frac12}{\frac12+\frac{\sqrt3}2}=\frac{\sqrt3+1}{\frac{\sqrt3+1}2}=(\sqrt3+1)\cdot\frac2{\sqrt3+1}=2$