## Elementary Geometry for College Students (5th Edition)

$x=50$ $\angle(1)=160^{\circ}$
Since $\angle(1)$ and $\angle(3)$ are equal, you can make an equation with their values as so: $3x+10=4x-40$ You can solve this to find the value for x: $3x+10=4x-40$ $(10+40)=(4x-3x)$ $50=x$ Now you can plug this value in to find the actual value of $\angle(1)$: $\angle(1)=(3x+10)^{\circ}$ $\angle(1)=(3(50)+10)^{\circ}$ $\angle(1)=(150+10)^{\circ}$ $\angle(1)=160^{\circ}$