Question

Demonstrati ca in orice triunghi
B=pi/6
C=pi/4
Este adevarata inegalitatea AB <2AC

Answer 1

Din teorema sinusurilor avem
$\displaystyle\frac{AC}{\sin B}=\frac{AB}{\sin C}\Rightarrow\frac{AC}{\frac{1}{2}}=\frac{AB}{\frac{\sqrt{2}}{2}}\Rightarrow AB=AC\sqrt{2}<2AC$

Answer 2

Teorema sinusului spune ca
$\frac{AB}{\sin{C}}=\frac{AC}{\sin{B}}$
Atunci in cazul nostru
$\frac{AB}{\sin{\frac{\pi}{4}}}=\frac{AC}{\sin{\frac{\pi}{6}}}\Rightarrow \frac{AB}{\frac{\sqrt{2}}{2}}=\frac{AC}{\frac{1}{2}}\Rightarrow \frac{AB}{\sqrt{2}}=AC\Rightarrow \frac{AB}{AC}=\sqrt{2}$
Atunci avem inegalitatea
$AB<2AC\Rightarrow \frac{AB}{AC}<2\Rightarrow \sqrt{2}<2\Rightarrow 2<4$ deci inegalitatea este adevarata