Answer:
[tex]x = 40 \\y = 35 \\\angle AEC = 55^\circ[/tex]
Step-by-step explanation:
Find x :
[tex]( 3x + 5)^{\circ} + ( x +15)^{ \circ} = 180 ^{\circ}[/tex] [tex][ \ straight \ line \ angles \ ][/tex]
[tex]3x + 5 + x + 15 = 180\\\\4x + 20 = 180 \\\\4x = 160 \\\\x = 40[/tex]
[tex]\angle AED \ = 3x + 5 = 125 ^{ \circ}[/tex]
[tex]\angle DEB = (x + 15 ) = 40 + 15 = 55^{ \circ}[/tex]
Find y :
[tex]( y + 20)^{ \circ} + (4y - 15)^{ \circ} = 180^{\circ}\\\\y + 20 + 4y - 15 = 180\\\\5y + 5 = 180\\\\5y = 175\\\\y = 35[/tex]
[tex]\angle AEC = ( y + 20 ) = 35 + 20 = 55^\circ\\\\\angle BEC = (4y - 15) = 4(35) -15 = 140 - 15 = 125^\circ[/tex]
