Question

Sendo A e B arcos de 1 ⁰ quadrante, tais que $\frac{sen A}{sen B} = \frac{1}{2}$ e $\frac{cos A}{cos B} = \frac{2\sqrt{10} }{5}$, determine tg A e tg B.

Explicar de forma didática, e quem vier com gracinha e respostas erradas será denúnciado. Ademais quem ajudar ganhará 25 pontos e 5 estrelas

Answer 1

Resposta:

$\text{tg}~\hat{\text{A}}=\dfrac{\sqrt{2}}{4}$ e $\text{tg}~\hat{\text{B}}=\dfrac{2\sqrt{5}}{5}$

Explicação passo-a-passo:

$\bullet~\dfrac{\text{sen}~\hat{\text{A}}}{\text{sen}~\hat{\text{B}}}=\dfrac{1}{2}~\Rightarrow~\dfrac{\text{sen}^2~\hat{\text{A}}}{\text{sen}^2~\hat{\text{B}}}=\dfrac{1}{4}$

Deste modo, $\text{sen}^2~\hat{\text{B}}=4\cdot\text{sen}^2~\hat{\text{A}}~~~(i)$

$\bullet~\dfrac{\text{cos}~\hat{\text{A}}}{\text{cos}~\hat{\text{B}}}=\dfrac{2\sqrt{10}}{5}~\Rightarrow~\dfrac{\text{cos}^2~\hat{\text{A}}}{\text{cos}^2~\hat{\text{B}}}=\dfrac{8}{5}$

Assim, $\text{cos}^2~\hat{\text{B}}=\dfrac{5\cdot\text{cos}^2~\text{A}}{8}~~~(ii)$

Somando (i) e (ii) membro a membro:

$\text{sen}^2~\hat{\text{B}}+\text{cos}^2~\hat{\text{B}}=4\cdot\text{sen}^2~\hat{\text{A}}+\dfrac{5\cdot\text{cos}^2~\text{A}}{8}$

Pela relação fundamental da trigonometria, temos que $\text{sen}^2~\hat{\text{A}}+\text{cos}^2~\hat{\text{A}}=1$ e $\text{sen}^2~\hat{\text{B}}+\text{cos}^2~\hat{\text{B}}=1$

Logo:

$4\cdot\text{sen}^2~\hat{\text{A}}+\dfrac{5\cdot\text{cos}^2~\text{A}}{8}=1$

Substituindo $\text{cos}^2~\hat{\text{A}}$ por $1-\text{sen}^2~\hat{\text{A}}$, segue que:

$4\cdot\text{sen}^2~\hat{\text{A}}+\dfrac{5\cdot(1-\text{sen}^2~\hat{\text{A}})}{8}=1$

$32\cdot\text{sen}^2~\hat{\text{A}}+5-5\cdot\text{sen}^2~\hat{\text{A}}=8$

$27\cdot\text{sen}^2~\hat{\text{A}}=3$

$\text{sen}~\hat{\text{A}}=\sqrt{\dfrac{1}{9}}$

$\text{sen}~\hat{\text{A}}=\dfrac{1}{3}$

$\text{cos}~\hat{\text{A}}=\sqrt{1-\left(\dfrac{1}{3}\right)^2}$

$\text{cos}~\hat{\text{A}}=\sqrt{\dfrac{8}{9}}$

$\text{cos}~\hat{\text{A}}=\dfrac{2\sqrt{2}}{3}$

Lembrando que $\hat{\text{A}}$ e $\hat{\text{B}}$ pertencem ao primeiro quadrante.

Substituindo esses valores nas equações dadas no enunciado:

$\bullet~\dfrac{\frac{1}{3}}{\text{sen}~\hat{\text{B}}}=\dfrac{1}{2}}~\Rightarrow~\text{sen}~\hat{\text{B}}=\dfrac{2}{3}$

$\bullet~\dfrac{\frac{2\sqrt{2}}{3}}{\text{cos}~\hat{\text{B}}}=\dfrac{2\sqrt{10}}{5}~\Rightarrow~\text{cos}~\hat{\text{B}}=\dfrac{\sqrt{5}}{3}$