Question

Calcule a medida dos lados AC e BC do triângulo. Depois determine a área do

triângulo ABC.

Answer 1

$\boxed{\begin{array}{l}\sf AC=x~~ BC=y\\\sf y\cdot sen(30^\circ)=3\sqrt{2} sen(45^\circ)\\\sf y\cdot\dfrac{1}{\diagup\!\!\!2}=3\sqrt{2}\cdot\dfrac{\sqrt{2}}{\diagup\!\!\!2}\\\\\sf y=3\sqrt{2^2}\\\sf y=3\cdot2\\\sf y=6\\\sf sen(105^\circ)=\dfrac{\sqrt{6}+\sqrt{2}}{4}\\\\\sf3\sqrt{2} sen(105^\circ)=x sen(30^\circ)\\\sf3\sqrt{2}\cdot\dfrac{\sqrt{6}+\sqrt{2}}{4}=x\cdot\dfrac{1}{2}\\\sf 4x=6\sqrt{2}(\sqrt{6}+\sqrt{2})\\\sf 4x=6\sqrt{12}+6\sqrt{2^2}\\\sf4x=6\cdot\sqrt{4\cdot3}+6\cdot2\\\sf 4x=6\cdot2\sqrt{3}+12\\\sf 4x=12\sqrt{3}+12\\\sf x=\dfrac{12\sqrt{3}+12}{4}\\\sf x=3\sqrt{3}+3\\\sf x=3\cdot(\sqrt{3}+1)\end{array}}$