Question

si los angulos exteriores de un polígono de 8 lados miden 2x,4x/3,x,5x/2,x/7,2x,x/12,3x/4 determinar el valor de cada angulo

Answer 1

Si los ángulos exteriores de un polígono de 8 lados miden 2x,4x/3,x,5x/2,x/7,2x,x/12,3x/4 determinar el valor de cada angulo

$2x\:+\frac{4}{3}x+x+\frac{5}{2}x+\frac{x}{7}+2x+\frac{x}{12}+\frac{3}{4}x=360$
$2\cdot \:2x+\frac{4}{3}x+x+\frac{5}{2}x+\frac{x}{7}+\frac{x}{12}+\frac{3}{4}x=360$
$4x+\frac{4}{3}x+x+\frac{5}{2}x+\frac{x}{7}+\frac{x}{12}+\frac{3}{4}x=360$
$336x+112x+84x+210x+12x+7x+63x=30240 824x=30240$
$x=\frac{3780}{103}$

Hallar angulos

Δ $2x=2\cdot \frac{3780}{103}=\frac{7560}{103}=73.398$

Δ $\frac{4}{3}x=\frac{4}{3}\cdot \frac{3780}{103}=\frac{5040}{103}=48.932$

Δ $x=\frac{3780}{103}=36.699$

Δ $\frac{5}{2}x = \frac{5}{2}\cdot \frac{3780}{103}=91.7475$

Δ $\frac{x}{7}=\frac{3780}{103}\div 7=\frac{540}{103}=5.2427$

Δ $2x=\frac{3780}{103}\cdot 2=\frac{7560}{103}=73.3980$

Δ $\frac{x}{12}=\frac{3780}{103}\div 12=\frac{315}{103}=3.05825$

Δ $\frac{3}{4}x=\frac{3}{4}\cdot \frac{3780}{103}=\frac{2835}{103}=27.524$