find the value of x,y and z.
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Answer:
in triangle ADC
A+D+C = 180
50⁰+90⁰+C = 180⁰
140⁰+C = 180⁰
C = 180⁰-140⁰
C = 40⁰
x = 90⁰
in triangle EBC
E+B+C = 180⁰
90⁰+y+40⁰ = 180⁰
150⁰+y = 180⁰
y = 180⁰-150⁰
y = 30⁰
E+D+O+C = 360⁰
90⁰+90⁰+O+40⁰ = 360⁰
180⁰+40⁰+O = 360⁰
220+O = 360⁰
O = 360⁰-220⁰
O = 140⁰
Z = O (V.O.A)
Z = 140⁰
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Step-by-step explanation:
in triangle AOE :
angle AOE = 180°-(50°+90°) = 40°
angle AOE = angle BOD = 40°
y+x+40 = 180 or x+y = 140° .....(1)
angle ABC = 2(angle EBC) = 2y
angle ABE = angle EBC = y
in triangle ABO:
y + z + angle ABO = 180
angle ABO = 180 - y - z. .....(2)
in triangle EBC :
angle ECB = 180-(90+y) = 90 - y
in quadrilateral OECD
EOD + ODC +DCE + CEO = 360
z + (180-x) + (90 - y) + 90 =360
z - x - y = 360 -180-90-90 = 0
z =x + y = 140. ( from eq 1)
in equation 3 :
angle ABC = 2y = z - y -50
2y + y = 140 - 50
3y = 90
y = 30
in equation 1 :
x+y =140
x = 140 -y = 140 - 30
x = 110