Questão 28 - Prova Geral Medicina - PUC Campinas 2026

I) $∆\mathrm{ABE}:$

$\mathrm{sen} 40° = \frac{5}{\mathrm{BE}} \Rightarrow \mathrm{BE} = \frac{5}{\mathrm{sen} 40°} \mathrm{cm}$

Além disso, tem-se $\mathrm{A}\hat{\mathrm{E}}\mathrm{B} = 50°.$

II) $∆\mathrm{CDE}:$

O triângulo é isósceles, logo $\mathrm{D}\hat{\mathrm{C}}\mathrm{E} = \mathrm{D}\hat{\mathrm{E}}\mathrm{C} = 45°.$

$\mathrm{sen} 45° = \frac{4}{\mathrm{CE}} \Rightarrow \frac{\sqrt{2}}{2} = \frac{4}{\mathrm{CE}} \Rightarrow \mathrm{CE} = 4\sqrt{2} \mathrm{cm}$

III) $∆\mathrm{BEC}:$

Como os pontos A, E e D estão alinhados, tem-se:

$50° + \mathrm{B}\hat{\mathrm{E}}\mathrm{C} + 45° = 180° \Rightarrow \mathrm{B}\hat{\mathrm{E}}\mathrm{C} = 85°.$

Pela lei dos cossenos:

${\mathrm{BC}}^2 = {\mathrm{BE}}^2 + {\mathrm{CE}}^2 - 2 · \mathrm{BE} · \mathrm{CE} · \cos 85°$

${\mathrm{BC}}^2 = {\left(\frac{5}{\mathrm{sen} 40°}\right)}^2 + {\left(4\sqrt{2}\right)}^2 - 2 · \frac{5}{\mathrm{sen} 40°} · 4\sqrt{2} · \cos 85°$

${\mathrm{BC}}^2 = \frac{25}{{\mathrm{sen}}^2 40°} + 32 - \frac{40\sqrt{2} · \cos 85°}{\mathrm{sen} 40°}$

${\mathrm{BC}}^2 = \frac{25 + 32 {\mathrm{sen}}^2 40° - 40\sqrt{2} · \mathrm{sen} 40° · \cos 85°}{{\mathrm{sen}}^2 40°}$

$\boxed{\mathrm{BC} = \frac{\sqrt{25 + 32 {\mathrm{sen}}^2 40° - 40\sqrt{2} · \mathrm{sen} 40° · \cos 85°}}{\mathrm{sen} 40°} \mathrm{cm}}$