Question

Sabendo que Z é o ponto medio do lado AF , o valor do ângulo k é :

a) 45°

b) 75°

c) 15°

d) 30°

e) 105°

Answer 1

Olá Lucas!

Resolvi a tarefa aplicando a Lei dos Senos e a Lei dos Cossenos, veja:

Considere $\mathsf{\overline{AZ} = \overline{ZF} = \frac{a}{2}}$, $\mathsf{\overline{AD} = b}$, $\mathsf{\overline{DF} = c}$ e $\mathsf{\overline{DZ} = h}$.

 Isto posto, considere o triângulo DFZ. Aplicando a Lei dos Senos nele teremos a seguinte proporção:

$\displaystyle \\ \mathsf{\frac{\frac{a}{2}}{\sin k} = \frac{h}{\sin 15^o}}$

Com efeito,

$\displaystyle \\ \mathsf{h = \frac{a}{2} \cdot \frac{\sin 15^o}{\sin k}}$


 Ademais, aplicamos a Lei dos Senos no triângulo ADF, veja:

$\displaystyle \\ \mathsf{\frac{a}{\sin 135^o} = \frac{b}{\sin 15^o} = \frac{c}{\sin 30^o} = \lambda, \ \forall \lambda \in \mathbb{N}^{\ast}}$


 Assim, a Lei dos Cossenos no $\Delta DFZ$ nos permite escrever:

$\displaystyle \\ \mathsf{h^2 = \left ( \frac{a}{2} \right )^2 + c^2 - 2 \cdot \frac{a}{2} \cdot c \cdot \cos 15^o}$

 A lei dos senos não foi aplicada à toa, faremos uso dela para que algumas incógnitas (h e c) desapareçam. Segue,

$\displaystyle \\ \mathsf{h^2 = \left ( \frac{a}{2} \right )^2 + c^2 - 2 \cdot \frac{a}{2} \cdot c \cdot \cos 15^o} \\\\\\ \mathsf{\left ( \frac{\sin 15^o}{\sin k} \cdot \frac{a}{2} \right )^2 = \frac{a^2}{4} + \left ( \frac{a \cdot \sin 30^o}{\sin 135^o} \right )^2 - a \cdot \left (\frac{a \cdot \sin 30^o}{\sin 135^o} \right ) \cdot \cos 15^o} \\\\\\ \mathsf{\frac{\sin^2 15^o}{\sin^2 k} \cdot \frac{a^2}{4} = \frac{a^2}{4} + \frac{a^2 \cdot \sin^2 30^o}{\sin^2 135^o} - \frac{a^2 \cdot \sin 30^o}{\sin 135^o} \cdot \cos 15^o \qquad \qquad \div (a^2} \\\\\\ \mathsf{\frac{\sin^2 15^o}{\sin^2 k} \cdot \frac{1}{4} = \frac{1}{4} + \frac{\sin^2 30^o}{\sin^2 135^o} - \frac{\sin 30^o}{\sin 135^o} \cdot \cos 15^o}$

$\displaystyle \\ \mathsf{\frac{\sin^2 15^o}{4 \cdot \sin^2 k} = \frac{1}{4} + \frac{\sin^2 30^o}{\sin^2 135^o} - \frac{\sin 30^o \cdot \cos 15^o}{\sin 135^o}} \\\\\\ \mathsf{\frac{1}{4} \cdot \left ( \frac{\sin 15^o}{\sin k} \right )^2 = \frac{1}{4} + \left ( \frac{\frac{1}{2}}{\frac{\sqrt{2}}{2}} \right )^2 - \frac{\frac{1}{2} \cdot \cos 15^o}{\frac{\sqrt{2}}{2}}} \\\\\\ \mathsf{\frac{1}{4} \cdot \left ( \frac{\sin 15^o}{\sin k} \right )^2 = \frac{1}{4} + \frac{1}{4} \cdot \frac{4}{2} - \frac{1}{2} \cdot \cos 15^o \cdot \frac{2}{\sqrt{2}}} \\\\\\ \mathsf{\frac{1}{4} \cdot \left ( \frac{\sin 15^o}{\sin k} \right )^2 = \frac{1}{4} + \frac{1}{2} - \frac{\cos 15^o}{\sqrt{2}}}$

$\displaystyle \\ \mathsf{\frac{1}{4} \cdot \left ( \frac{\sin 15^o}{\sin k} \right )^2 = \frac{\sqrt{2} + 2\sqrt{2} - 4 \cdot \cos 15^o}{4\sqrt{2}}} \\\\\\ \mathsf{\left ( \frac{\sin 15^o}{\sin k} \right )^2 = \frac{3\sqrt{2} - 4 \cdot \cos 15^o}{\sqrt{2}}} \\\\\\ \mathsf{\sin^2 k \cdot \left ( 3\sqrt{2} - 4 \cdot \cos 15^o \right ) = \left ( \sin 15^o \right )^2 \cdot \sqrt{2}} \\\\\\ \mathsf{\sin^2 k \cdot \left [ 3\sqrt{2} - 4 \cdot \cos (45^o - 30^o) \right ] = \left [ \sin (45^o - 30^o) \right ]^2 \cdot \sqrt{2}}$
 
 Obs1.: Cos

$\displaystyle \\ \mathsf{\cos 15^o = \cos (45^o - 30^o)} \\\\ \mathsf{\cos 15^o = \cos 45^o \cdot \cos 30^o + \sin 45^o \cdot \sin 30^o} \\\\ \mathsf{\cos 15^o = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}} \\\\ \mathsf{\cos 15^o = \frac{\sqrt{6} + \sqrt{2}}{4}}$

Obs2.: sin

$\displaystyle \\ \mathsf{\sin 15^o = \sin (45^o - 30^o)} \\\\ \mathsf{\sin 15^o = \sin 45^o \cdot \cos 30^o - \sin 30^o \cdot \cos 45^o} \\\\ \mathsf{\sin 15^o = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2}} \\\\ \mathsf{\sin 15^o = \frac{\sqrt{6} - \sqrt{2}}{4}}$


 Daí,

$\displaystyle \\ \mathsf{\sin^2 k \cdot \left ( 3\sqrt{2} - 4 \cdot \cos 15^o \right ) = \left ( \sin 15^o \right )^2 \cdot \sqrt{2}} \\\\\\ \mathsf{\sin^2 k \cdot \left ( 3\sqrt{2} - 4 \cdot \frac{\sqrt{6} + \sqrt{2}}{4} \right ) = \left ( \frac{\sqrt{6} - \sqrt{2}}{4} \right )^2 \cdot \sqrt{2}} \\\\\\ \mathsf{\sin^2 k \cdot (3\sqrt{2} - \sqrt{6} - \sqrt{2}) = \frac{6 - 2\sqrt{12} + 2}{16} \cdot \sqrt{2}} \\\\\\ \mathsf{\sin^2 k \cdot (2\sqrt{2} - \sqrt{6}) = \frac{8 - 4\sqrt{3}}{16} \cdot \sqrt{2}}$

$\displaystyle \\ \mathsf{\sqrt{2} \cdot (2 - \sqrt{3}) \cdot \sin^2 k = \frac{4(2 - \sqrt{3})}{16} \cdot \sqrt{2} \qquad \qquad \div[\sqrt{2}(2 - \sqrt{3})} \\\\\\ \mathsf{\sin^2 k = \frac{4}{16}} \\\\\\ \mathsf{\sin k = \sqrt{\frac{4}{16}}} \\\\\\ \mathsf{\sin k = \frac{2}{4}} \\\\\\ \mathsf{\sin k = \frac{1}{2}} \\\\\\ \boxed{\boxed{\mathsf{k = 30^o}}}$

 Espero ter ajudado!!

Bons estudos.

Qualquer dúvida, comente!!