Question

[Lei dos Cossenos] Num triângulo cujo a medida AB=10cm, CB=20cm, BA=x, calcule a medida de BA sabendo que o ângulo  é igual a 89°.

Answer 1

$Lei\ de\ forma\c{c}\~ao\ dos\ cossenos:\\\\ \overline{AB}^2=\overline{BC}^2+\overline{AC}^2-2.\overline{BC}.\overline{AC}.\cos (\overline{AB})^o\\\\ \overline{BC}^2=\overline{AB}^2+\overline{AC}^2-2.\overline{AB}.\overline{AC}.\cos (\overline{BC})^o\\\\ \overline{AC}^2=\overline{AB}^2+\overline{BC}^2-2.\overline{AB}.\overline{BC}.\cos (\overline{AC})^o\\\\\\ Temos:\\\\ \overline{AB}=x\\\\ \overline{BC}=20\ cm\\\\ \overline{AC}=10\ cm\\\\ \^A=(\overline{BC})^o=89^o$


Como temos o valor de Â, então utilizaremos:

$\overline{BC}^2=\overline{AB}^2+\overline{AC}^2-2.\overline{AB}.\overline{AC}.\cos \^A\\\\ \cos 89^o\approx 0,01745\\\\\\ Resolvendo:\\\\ 20^2=x^2+10^2-2.x.(10).(0,01745)\\\\ 400=x^2+100-0,349x\\\\ x^2-0,349x-300=0$


$Aplicando\ Bh\'askara:\\\\ x=\dfrac{-b\pm\sqrt{b^2-4.a.c}}{2.a}\\\\ x=\dfrac{0,349\pm\sqrt{0,121801+1.200}}{2}\\\\ x \approx \dfrac{0,349\pm34,64}{2}\ (\ utilizaremos\ apenas\ o\ valor\ positivo\ )\\\\ x\approx\dfrac{34,989}{2}\\\\ \boxed{x\approx17,4945\ cm}$


$\boxed{A\ medida\ \overline{AB}\ vale\ aproximadamente\ 17,50\ cm}$


Bons estudos!