Cap. 5 - Triângulos II ⇒ Solucionário:Racso - Cap V - Problemas de Geometria y Como Resolverlos - I Edição - Ex:40 Tópico resolvido

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Problema Proposto (Último do capítulo V)
40 - Calcular x na figura

Resposta

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D) 18^o30'

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Por trigonometria

[tex3]\angle ADC=45-2x\\
\angle ABC=45-x\\
\frac{AC}{\sin (45-2x)}=\frac{AD}{\sin 135}\\
Lei~Seno: \triangle ACD: \frac{AC}{\sin (45-2x)}=\frac{AD}{\sin 135}(I) \\Lei~Seno: \triangle ABD:\frac{AB}{\sin (x)}=\frac{AD}{\sin 135}(II)\\
De(I)e(II):\frac{AC}{\sin (45-2x)}=\frac{AB}{\sin (x)}\rightarrow \frac{AC}{AB}=\frac{\sin (45-2x)}{\sin (x)}\\
\triangle ABC:sin (45-x)=\frac{AC}{AB}\rightarrow \frac{\sin (45-2x)}{\sin (x)}=\sin (45-x)\\
\text{Usando identidades de soma / diferença }\\\frac{\sqrt{2}}{2}(\cos (2x)-\sin (2x))=\frac{\sqrt{2}}{2}(\sin (x)\cos (x)-\sin^2 x)\rightarrow \\
\cos^2 (x)-\sin^2 (x)-2\sin (x)\cos (x)=\sin (x)\cos (x)-\sin^2 (x)\rightarrow \\
\cos^2 (x)=3\sin (x)\cos (x)\rightarrow se~cos(x) = 0 \rightarrow x = \frac{\pi}{2}\\
mas~x < \frac{\pi}{2}\therefore \cos (x)=3\sin (x)\rightarrow tgx = \frac{1}{3}\therefore \boxed{\color{red}x =\approx18,4}
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Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\angle ADC=45-2x\\
\angle ABC=45-x\\
\frac{AC}{\sin (45-2x)}=\frac{AD}{\sin 135}\\
Lei~Seno: \triangle ACD: \frac{AC}{\sin (45-2x)}=\frac{AD}{\sin 135}(I) \\Lei~Seno: \triangle ABD:\frac{AB}{\sin (x)}=\frac{AD}{\sin 135}(II)\\
De(I)e(II):\frac{AC}{\sin (45-2x)}=\frac{AB}{\sin (x)}\rightarrow \frac{AC}{AB}=\frac{\sin (45-2x)}{\sin (x)}\\
\triangle ABC:sin (45-x)=\frac{AC}{AB}\rightarrow \frac{\sin (45-2x)}{\sin (x)}=\sin (45-x)\\
\text{Usando identidades de soma / diferença }\\\frac{\sqrt{2}}{2}(\cos (2x)-\sin (2x))=\frac{\sqrt{2}}{2}(\sin (x)\cos (x)-\sin^2 x)\rightarrow \\
\cos^2 (x)-\sin^2 (x)-2\sin (x)\cos (x)=\sin (x)\cos (x)-\sin^2 (x)\rightarrow \\
\cos^2 (x)=3\sin (x)\cos (x)\rightarrow se~cos(x) = 0 \rightarrow x = \frac{\pi}{2}\\
mas~x < \frac{\pi}{2}\therefore \cos (x)=3\sin (x)\rightarrow tgx = \frac{1}{3}\therefore \boxed{\color{red}x =\approx18,4}
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(Solução: Alan Abraham)

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[tex3]\mathsf{
BC=a,AC=b.
BH \perp AD\\
△ABC≅△ABH \implies ∠ABC=∠BAH=45∘−x (AB~hipotenusa~comum)\\
\therefore AH=BC=a\\
BH=AC=b\\
Analogamente: △DBE≅△BDH. \implies DE=BH=b\\
DH=BE ~mas BE=CF=DF=a+b \implies\\
DF=a+b,AF=a+2b,AD=2a+b\\
T.Pit: (a+b)^2+(a+2b)^2=(2a+b)^2⟹(a+b)(a−2b)=0\\
\therefore a=2b \implies △ADF (notável) 3:4:5
\therefore \boxed{\color{red} 2x≈37^∘\implies x = 18^o30'}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{
BC=a,AC=b.
BH \perp AD\\
△ABC≅△ABH \implies ∠ABC=∠BAH=45∘−x (AB~hipotenusa~comum)\\
\therefore AH=BC=a\\
BH=AC=b\\
Analogamente: △DBE≅△BDH. \implies DE=BH=b\\
DH=BE ~mas BE=CF=DF=a+b \implies\\
DF=a+b,AF=a+2b,AD=2a+b\\
T.Pit: (a+b)^2+(a+2b)^2=(2a+b)^2⟹(a+b)(a−2b)=0\\
\therefore a=2b \implies △ADF (notável) 3:4:5
\therefore \boxed{\color{red} 2x≈37^∘\implies x = 18^o30'}}[/tex3]

(Solução:MathLover)