Answer:
Explanation:
1.
∠ACB=6 x°
∠DCE=30°
6x=30°-----------------divide both sides by 6
x=30°/6= 5°
∠BCE=180°-30°=150°----------------sum of angles on a straight line
2.
∠ABD=(4+5x)°
∠CBD=(x+2)°
(4+5x)°+(x+2)°=180°-----------sum of angles on a straight line
4+5x+x+2=180°----------------collect like terms
6x+6=180°
6x=180°-6°
6x=174°------------divide by 6 both sides
x=174°÷6= 29°
∠ABD=4+5x=4+(5*29)= 4+145=149°
∠CBD= x+2=29°+2°=31°
3.
BE bisects ∠ABE=(3x+1)°
m∠DBA=(8x-14)°
1/2 (m∠DBA)° = m∠ABE
1/2(8x-14)°=(3x+1)°
(4x-7)°=(3x+1)°
4x-7=3x+1
4x-3x=7+1-------------collecting like terms
x=8°
m∠ABE=(3x+1)°=(3*8+1)=24+1=25°
m∠DBA=(8x-14)°= (8*8-14)°=(64-14)=50°
4.
∠ADE=∠CDG
50+3x-y=x+2x-16------------collect like terms
50+16-y=x+2x-3x
66-y=3x-3x
66°=y
∠ADC +∠ADB+∠BDE
90°+50°+(3x-y)°=180°
140°+(3x-y)°=180°
3x-66°=180°-140°
3x=40°+66°
3x=106°-----------divide both sides by 3
x=106°÷3= 35.33°
∠FDG=(2x-16)°= (2×35.33° - 16° )= 54.67°
∠BDE = (3x-y)°= (3×35.33°-66°)= 105.99-66=39.99°
5.
∠ABD+∠DBC=90°
(6x+4)°+32°=90°
(6x+4)°=90°-32°
6x+4=58°
6x=58°-4°
6x=54°
x=54°÷6=9°
∠ABD= (6x+4)°+32°
∠ABD= (6×9 +4)°+32°
∠ABD= (54°+4°) + 32°
∠ABD=58°+32°=90°
6.
∠AED=∠CEB
(3x+5)°=(4y-15)°------------------form equation of equality
3x+5=4y-15
5+15=4y-3x
20=4y-3x------------------------------(1)
∠AEC=∠DEB
(y+20)°=(x+15)°
y+20=x+15
20-15=x-y
5°=x-y
5+y=x----------------------------(2)
Use equation (2) in equation (1)
20=4y-3x
20=4y-3(5+y)
20=4y-15-3y
20+15=4y-3y
35°=y
Solve for x
x=5+y=5°+35°=40°
∠AED=(3x+5)°=(3×40 +5 )=120+5=125°
∠AEC= (y+20)°= 35° + 20° =55°