Question

Calcule a base de um triângulo isósceles em que os lados iguais medem 20 metros e os ângulos congruentes medem 72

Answer 1

Boa noite Vyttor

72 + 72 + α = 180

α + 144 = 180
α = 180 - 144 = 36°

Lei dos cossenos

b² = 20² + 20² - 2*20*20*cos(36°)
b² = 400 + 400 - 800*0.81
b² = 800*(1 - 0.81)
b² = 800*0.19 = 152

b = √152 = 12.36 m

Answer 2

$\textbf{Aplicando a lei dos cossenos:}\\\\ \mathrm{b^2=20^2+20^2-2.20.20.\cos{\gamma}}\\ \mathrm{b^2=400+400-2.400.\cos{\gamma}}\\ \mathrm{b^2=800-800\cos{\gamma}\ \to\ b^2=800(1-\cos{\gamma})}\\ \mathrm{b^2=800.2\sin^2{\bigg(\dfrac{\gamma}{2}\bigg)}\ \to\ b^2=1600\sin^2{\bigg(\dfrac{\gamma}{2}\bigg)}}\\\\ \mathrm{b=\sqrt{1600\sin^2{\bigg(\dfrac{\gamma}{2}\bigg)}}\ \to\ b=40\sin{\bigg(\dfrac{\gamma}{2}\bigg)}}$

$\textbf{Aplicando a lei dos senos:}\\\\ \mathrm{\dfrac{20}{\sin{2\gamma}}=\dfrac{b}{\sin{\gamma}}\ \to\ \dfrac{20}{2\sin{\gamma}\cos{\gamma}}=\dfrac{b}{\sin{\gamma}}}\\\\ \mathrm{b=\dfrac{10}{\cos{\gamma}}\ \to\ \dfrac{10}{\cos{\gamma}}=40\sin{\bigg(\dfrac{\gamma}{2}\bigg)}\ \to\ \dfrac{1}{4}=\cos{\gamma}\sqrt{\dfrac{1-\cos{\gamma}}{2}}}\\\\ \mathrm{\dfrac{1}{16}=\cos^2{\gamma}\bigg(\dfrac{1-\cos{\gamma}}{2}\bigg)\ \to\ \dfrac{1}{8}=\cos^2{\gamma}(1-\cos{\gamma})}\\\\ \mathrm{0=\cos^2{\gamma}-\cos^3{\gamma}-\dfrac{1}{8}\ \to\ 8\cos^3{\gamma}-8\cos^2{\gamma}+1=0}\\\\ \mathrm{*\ Para\ \cos{\gamma}=\dfrac{1}{2}\ \to\ 8.\dfrac{1}{8}-8.\dfrac{1}{4}+1=1-2+1=0}\\\\ \mathrm{*\ Fatorando\ a\ express\~ao:}\\\\ \mathrm{(2\cos{\gamma}-1)\bigg(\dfrac{8\cos^3{\gamma}-8\cos^2{\gamma}+1}{2\cos{\gamma}-1}\bigg)=}\\\\ \mathrm{=(2\cos{\gamma}-1)(4\cos{\gamma}^2-2\cos{\gamma}-1)=0}$
$\mathrm{4\cos^2{\gamma}-2\cos{\gamma}-1=0\ \to\ a=4\ \|\ b=-2\ \|\ c=-1}\\\\ \mathrm{\cos{\gamma}=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{2\pm\sqrt{(-2)^2-4.4.(-1)}}{2.4}=}\\\\ \mathrm{=\dfrac{2\pm\sqrt{4+16}}{8}=\dfrac{2\pm\sqrt{20}}{8}=\dfrac{2\pm2\sqrt{5}}{8}=\dfrac{1\pm\sqrt{5}}{4}}\\\\ \mathrm{*\ \cos{\gamma}=\dfrac{1}{2}\ e\ \cos{\gamma}=\dfrac{1-\sqrt{5}}{2}\ \to\ n\~ao\ conv\'em}\\\\ \mathrm{*\ \cos{\gamma}=\dfrac{1+\sqrt{5}}{4}}$

$\textbf{Voltando \`a lei dos senos:}\\\\ \mathrm{b=\dfrac{10}{\cos{\gamma}}=\dfrac{10}{\frac{1+\sqrt{5}}{4}}=\dfrac{40}{1+\sqrt{5}}}\\\\ \mathrm{b=\dfrac{40}{\sqrt{5}+1}.\dfrac{\sqrt{5}-1}{\sqrt{5}-1}=\dfrac{40(\sqrt{5}-1)}{(\sqrt{5})^2-1^2}}\\\\ \mathrm{b=\dfrac{40(\sqrt{5}-1)}{5-1}=\dfrac{40(\sqrt{5}-1)}{4}}\\\\ \boxed{\mathbf{b=10(\sqrt{5}-1)\approx12,36\ m}}$