Question

Ajuda nesta questão de trigonometria [URGENTE]

Answer 1

Olhando pro triângulo retângulo ABD:

$sen~60\º=BA/AD\\sen~60\º=BA/20\\\sqrt{3}/2=BA/20\\\sqrt{3}/1=BA/10\\10*\sqrt{3}=BA\\BA=10\sqrt{3}~cm$

Olhando pro triângulo retângulo ABC:

$sen~45\º=BA/BC\\sen~45\º=10\sqrt{3}/x\\\sqrt{2}/2=10\sqrt{3}/x\\x*\sqrt{2}=2*10\sqrt{3}\\x=20\sqrt{3}/\sqrt{2}\\x=20\sqrt{3}*\sqrt{2}/(\sqrt{2}*\sqrt{2})\\x=20\sqrt{3*2}/2\\x=10\sqrt{6}~cm$