Estude! Com Prof. Caju

Em um triângulo ABC, AB=7 e BC=8; Prolongando CA e BC marca os pontos R e M.Se 16(RA)=11(RC)=176, RE=14 e m < ABC=m < CRE (E é ponto exterior relativo a AC do triângulo ABC, calcule m < ECM.
a)60°
b)30°
c)[tex3]\frac{37°}{2}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{37°}{2}[/tex3]

d)[tex3]\frac{53°}{2}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{53°}{2}[/tex3]

e)53°

botelho,

[tex3]RA = \frac{176}{16}=11:RC = \frac{176}{11}=16 \implies AC = 5
\\
\frac{RE}{AB} =\frac{14}{7}=2 =\frac{RC}{BC} =\frac{16}{8}
\\\triangle ECR \sim \triangle ACB\\
\therefore \frac{EC}{AC} = 2 \implies \frac{EC}{5}=2 \therefore EC = 10\\
\triangle ECR:\angle ECR=\theta\\
T.Cossenos: 14^2 = 10^2+!6^2-2.10.16cos\theta \implies cos\theta = \frac{1}{2} \therefore \theta = 60^o \\
\angle ACB \cong \angle ECR=60^o \implies \angle ECM =180^o -120^o = \boxed{60^o }


[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]RA = \frac{176}{16}=11:RC = \frac{176}{11}=16 \implies AC = 5
\\
\frac{RE}{AB} =\frac{14}{7}=2 =\frac{RC}{BC} =\frac{16}{8}
\\\triangle ECR \sim \triangle ACB\\
\therefore \frac{EC}{AC} = 2 \implies \frac{EC}{5}=2 \therefore EC = 10\\
\triangle ECR:\angle ECR=\theta\\
T.Cossenos: 14^2 = 10^2+!6^2-2.10.16cos\theta \implies cos\theta = \frac{1}{2} \therefore \theta = 60^o \\
\angle ACB \cong \angle ECR=60^o \implies \angle ECM =180^o -120^o = \boxed{60^o }


[/tex3]