TRIGONOMETRIA 19. Determine o valor de x nos triângulos a seguir Respostas incompletas, com brincade.... Pergunta de ideia deSubGui

December 2019 1 183 Report

TRIGONOMETRIA

19. Determine o valor de x nos triângulos a seguir

Respostas incompletas, com brincadeiras ou sem sentido serão desconsideradas

Lista de comentários

FibonacciTH ๏ Lei dos cossenos (Auxilio Figura 01):

$\hookrightarrow \mathsf{a^2=b^2+c^2-2bc\cdot cos\left(\alpha \right)}\\\hookrightarrow \mathsf{b^2=a^2+c^2-2ac\cdot cos\left(\beta \right)}\\\hookrightarrow \mathsf{c^2=a^2+b^2-2ab\cdot cos\left(\theta \right)}$

๏ Lei do senos (Auxilio Figura 02):

$\mathsf{\dfrac{a}{sen\left(a\right)}=\dfrac{b}{sen\left(b\right)}=\dfrac{c}{sen\left(c\right)}}$
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a) Triangulo de lados 25 cm , 55 cm e x cm:

$\mathsf{25^2=x^2+55^2-\left[2\cdot x\cdot 55\cdot cos\left(30^{\circ}\right)\right]}\\\\\mathsf{625=x^2+3025-\left[2\cdot x\cdot 55\cdot cos\left(30\right)\right]}\\\\\mathsf{625=x^2+3025-\left[2\cdot x\cdot 55\cdot \dfrac{\sqrt{3}}{2}\right]}\\\\\mathsf{625=x^2+3025-\left(55x\cdot \sqrt{3}\right)}\\\\\mathsf{x^2-55x\sqrt{3}+3025-625=0}\\\\\mathsf{x^2-55x\sqrt{3}+2400=0}\\\\\\\mathsf{a=1;\:b=55\sqrt{3};\:c=2400}$

$\mathsf{\Delta =b^2-4ac}\\\mathsf{\Delta =\left(55\sqrt{3}\right)^2-\left(4\cdot 1\cdot 2400\right)}\\\mathsf{\Delta =\left(3025\cdot 3\right)-9600}\\\mathsf{\Delta =9075-9600}\\\mathsf{\Delta =-525}\\\\\\\boxed{\boxed{\mathsf{S=\left\{\phi ,\:\forall x\in \mathbb{R}\right\}}}}\:\: \checkmark$

b) Triangulo de lados x cm e 8√3 cm. Aplicando a lei dos senos:

$\mathsf{\dfrac{8\sqrt{3}}{sen\left(60^{\circ }\right)}=\dfrac{x}{sen\left(45^{\circ }\right)}}\\\\\\\mathsf{x=\dfrac{8\sqrt{3}\cdot sen\left(45^{\circ}\right)}{sen\left(60^{\circ }\right)}}\\\\\\\mathsf{x=\dfrac{8\sqrt{3}\cdot \dfrac{\sqrt{2}}{2}}{\dfrac{\sqrt{3}}{2}}}\\\\\\\mathsf{x=\dfrac{8\sqrt{3}\cdot \sqrt{2}\cdot 2}{2\sqrt{3}}}\\\\\\\boxed{\boxed{\mathsf{x=8\sqrt{2}\:cm}}}\:\: \checkmark$

c) Triangulo de lados 3,5 cm , 2 cm e x cm. Aplicando a Lei dos cossenos:

$\mathsf{x^2=\left(3,5\right)^2+2^2-\left[2\cdot 3,5\cdot 2\cdot cos\left(120^{\circ }\right)\right]}\\\\\\\mathsf{x^2=\left(\dfrac{7}{2}\right)^2+4+\left[2\cdot \dfrac{7}{2}\cdot 2\cdot cos\left(60^{\circ }\right)\right]}\\\\\\\mathsf{x^2=\dfrac{49}{4}+4+\left[14\cdot \dfrac{1}{2}\right]}\\\\\\\mathsf{x^2=\dfrac{49}{4}+4+7}\\\\\\\mathsf{x^2=\dfrac{49}{4}+11}\\\\\\\mathsf{x^2=\dfrac{49}{4}+\dfrac{11\cdot 4}{4}}\\\\\\\mathsf{x^2=\dfrac{49+44}{4}}\\\\\\\mathsf{x^2=\dfrac{93}{4}}$
$\\\\\\\mathsf{x=\sqrt{\dfrac{93}{4}}}\\\\\\\mathsf{x=\dfrac{\sqrt{93}}{\sqrt{4}}}\\\\\\\boxed{\boxed{\mathsf{x=\dfrac{\sqrt{93}}{2}\:cm}}}\:\: \checkmark$

d) Triangulo de lados x cm e 8 cm. Aplicando a lei dos senos:

$\mathsf{\dfrac{8}{sen\left(45^{\circ }\right)}=\dfrac{x}{sen\left(30^{\circ }\right)}}\\\\\\\mathsf{x=\dfrac{8\cdot sen\left(30^{\circ }\right)}{sen\left(45^{\circ }\right)}}\\\\\\\mathsf{x=\dfrac{8\cdot \dfrac{1}{2}}{\dfrac{\sqrt{2}}{2}}}\\\\\\\mathsf{x=\dfrac{8\cdot \:2}{2\sqrt{2}}}\\\\\\\mathsf{x=\dfrac{8}{\sqrt{2}}}\\\\\\\mathsf{x=\dfrac{8\cdot \sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}}\\\\\\\mathsf{x=\dfrac{8\sqrt{2}}{2}}\\\\\\\boxed{\boxed{\mathsf{x=4\sqrt{2}\:cm}}}\:\: \checkmark$

e) Triangulo de lados x cm e 37 cm.

Considerando:

๏ $\mathsf{sen\left(62^{\circ }\right)\approx 0,88}$
๏ $\mathsf{sen\left(48^{\circ }\right)\approx 0,74}$

Aplicando a formula da "Lei dos senos":

$\mathsf{\dfrac{x}{sen\left(62^{\circ }\right)}=\dfrac{37}{sen\left(48^{\circ }\right)}}\\\\\\\mathsf{x=\dfrac{37}{sen\left(48^{\circ }\right)}\cdot sen\left(62^{\circ }\right)}\\\\\\\mathsf{x=\dfrac{37\cdot 0,88}{0,74}}\\\\\\\boxed{\boxed{\mathsf{x=44\:cm}}}\:\:\checkmark$

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