Question

Me ajudem nesta questão de matemática ?! U-R-G-E-N-T-E

Answer 1

$\bullet~~\mathrm{m}(A\widehat{O}C)=45^{\circ}\\\\ \bullet~~C\widehat{O}D\cong D\widehat{O}E$


Lembrando da transformação de graus para minutos:

$1^{\circ}=60'$

(um grau é igual a sessenta minutos)


a) $\mathrm{m}(A\widehat{O}C)=45^{\circ}$

$x+x=45^{\circ}\\\\ 2x=45^{\circ}\\\\ 2x=44^{\circ}+1^{\circ}\\\\ 2x=44^{\circ}+60'\\\\ x=\dfrac{44^{\circ}+60'}{2}\\\\ x=22^{\circ}+30'\\\\ \boxed{\begin{array}{c}x=22^{\circ}\,30' \end{array}}$


b) $\mathrm{m}(A\widehat{O}D)=\mathrm{m}(A\widehat{O}B)+\mathrm{m}(B\widehat{O}C)+\mathrm{m}(C\widehat{O}D)$

$\mathrm{m}(A\widehat{O}D)=x+x+(3x-12^{\circ}\,30')\\\\ \mathrm{m}(A\widehat{O}D)=2x+3x-12^{\circ}\,30'\\\\ \mathrm{m}(A\widehat{O}D)=5x-12^{\circ}\,30'\\\\ \mathrm{m}(A\widehat{O}D)=5\cdot (22^{\circ}\,30')-12^{\circ}\,30'\\\\ \mathrm{m}(A\widehat{O}D)=5\cdot (22^{\circ}+30')-(12^{\circ}+30')\\\\ \mathrm{m}(A\widehat{O}D)=(5\cdot 22^{\circ}+5\cdot 30')-(12^{\circ}+30')\\\\ \mathrm{m}(A\widehat{O}D)=(110^{\circ}+150')-(12^{\circ}+30')\\\\ \mathrm{m}(A\widehat{O}D)=110^{\circ}+150'-12^{\circ}-30'$


Operamos graus com graus, e minutos com minutos:

$\mathrm{m}(A\widehat{O}D)=(110^{\circ}-12^{\circ})+(150'-30')\\\\ \mathrm{m}(A\widehat{O}D)=98^{\circ}+120'\\\\ \mathrm{m}(A\widehat{O}D)=98^{\circ}+2\cdot 60'\\\\ \mathrm{m}(A\widehat{O}D)=98^{\circ}+2\cdot 1^{\circ}\\\\ \mathrm{m}(A\widehat{O}D)=98^{\circ}+2^{\circ}\\\\ \boxed{\begin{array}{c}\mathrm{m}(A\widehat{O}D)=100^{\circ} \end{array}}$


c) $\mathrm{m}(D\widehat{O}E)=\mathrm{m}(C\widehat{O}D)$

(porque os dois ângulos são congruentes entre si)

$\mathrm{m}(D\widehat{O}E)=3x-12^{\circ}\,30'\\\\ \mathrm{m}(D\widehat{O}E)=3\cdot (22^{\circ}\,30')-12^{\circ}\,30'\\\\ \mathrm{m}(D\widehat{O}E)=3\cdot (22^{\circ}+30')-(12^{\circ}+30')\\\\ \mathrm{m}(D\widehat{O}E)=(3\cdot 22^{\circ}+3\cdot 30')-(12^{\circ}+30')\\\\ \mathrm{m}(D\widehat{O}E)=(66^{\circ}+90')-(12^{\circ}+30')\\\\ \mathrm{m}(D\widehat{O}E)=66^{\circ}+90'-12^{\circ}-30'$


Graus com graus, e minutos com minutos:

$\mathrm{m}(D\widehat{O}E)=(66^{\circ}-12^{\circ})+(90'-30')\\\\ \mathrm{m}(D\widehat{O}E)=54^{\circ}+60'\\\\ \mathrm{m}(D\widehat{O}E)=54^{\circ}+1^{\circ}\\\\ \boxed{\begin{array}{c} \mathrm{m}(D\widehat{O}E)=55^{\circ} \end{array}}$