Question

Num triângulo ABC o ângulo A = 30° é oposto ao lado a = 15 cm. Sabendo que SenB +
SenC = 4/3. Calcule o perímetro do triângulo.

Answer 1

Lei dos senos :

$\displaystyle\sf \frac{a}{Sen(A)}=\frac{b}{Sen(B)}=\frac{c}{Sen(C)} \\\\\\ temos : \\\\ \left\{ \begin{array} \displaystyle\sf a = 15 \ cm \ \\\\ \displaystyle\sf sen(A) = sen(30)=\frac{1}{2}\ \\\\ \displaystyle\sf Sen(B)+Sen(C)=\frac{4}{3}\end{array} \right$

$\displaystyle \sf Da{\'i}} : \\\\ \frac{15}{\displaystyle \frac{1}{2}} = \frac{b}{Sen(B)}=\frac{c}{Sen(C)} \\\\\\ \frac{b}{Sen(B)}=\frac{c}{Sen(C)} = 30 \\\\\\\ \underline{isolando \ os\ lados } : \\\\ \left\{\begin{array}{}\displaystyle\sf \frac{b}{Sen(B)} = 30 \to b = 30\cdot Sen(B) \\\\\\ \displaystyle\sf \frac{c}{Sen(C)} = 30 \to c = 30\cdot Sen(C)\end{array} \right$

A questão pede o perímetro(2P) portanto :

$\displaystyle \sf 2P = a+b+c \\\\ 2P = 15+30\cdot Sen(B)+30\cdot Sen(C) \\\\ 2P=15+30\cdot (Sen(B)+Sen(C)) \\\\ 2P = 15+30\cdot \frac{4}{3} \\\\ 2P = 15 +10\cdot 4 \\\\ 2P = 15+40 \\\\ \huge\boxed{\sf 2P=55 \ cm }\checkmark$