Question

Na figura abaixo, BQA= 45 °e DPC= 15°. Calcule os valores dos arcos AB e CD.

Answer 1

Explicação passo-a-passo:

Sejam $\sf arco~AB=x$ e $\sf arco~CD=y$

• Ângulo excêntrico interno

$\sf x=\dfrac{arco~AE+arco~BD}{2}$

Temos:

$\sf B\hat{Q}A=\dfrac{arco~AB+arco~CD}{2}$

$\sf 45^{\circ}=\dfrac{x+y}{2}$

$\sf x+y=2\cdot45^{\circ}$

$\sf x+y=90^{\circ}$

• Ângulo excêntrico exterior

$\sf \theta=\dfrac{\alpha-\beta}{2}$

$\sf D\hat{P}C=\dfrac{arco~AB-arco~BD}{2}$

$\sf 15^{\circ}=\dfrac{x-y}{2}$

$\sf x-y=2\cdot15^{\circ}$

$\sf x-y=30^{\circ}$

Podemos montar o sistema:

$\sf \begin{cases} \sf x+y=90^{\circ} \\ \sf x-y=30^{\circ} \end{cases}$

Somando as equações:

$\sf x+x+y-y=90^{\circ}+30^{\circ}$

$\sf 2x=120^{\circ}$

$\sf x=\dfrac{120^{\circ}}{2}$

$\sf x=60^{\circ}$

Substituindo na primeira equação:

$\sf x+y=90^{\circ}$

$\sf 60^{\circ}+y=90^{\circ}$

$\sf y=90^{\circ}-60^{\circ}$

$\sf y=30^{\circ}$

Assim:

• $\sf \large\red{arco~AB=60^{\circ}}$

• $\sf \large\red{arco~CD=30^{\circ}}$