In the adjacent figure x:y=11:19.if AD//BE find the angle DCE
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9
180-(x+y) will be the answer of angle DCE
ErNilesh:
angle of a will be 180-x-y in ADB that's why the angle of bcd will be X+y in quadrant adcb so angle of DCE WILL BE 180-(X+Y) BECOS THE ANGLE BCE IS 180
Answer is 65°
Not 180°
how to get value of x
then what is answer
The ques should be AB⊥BD
Solving them
let x and y be 11a and 19a
then in ΔABD
11a+19a+90°=180° (angle sum property)
30a=90°
a=90°/30 = a=30°
∠x=33° (11a)
∠y=57°(19a)
as AB║BD
Then,∠ADC=∠DCE
∠DCE=65° (32+33)
:) Hope this helps!!!
Solving them
let x and y be 11a and 19a
then in ΔABD
11a+19a+90°=180° (angle sum property)
30a=90°
a=90°/30 = a=30°
∠x=33° (11a)
∠y=57°(19a)
as AB║BD
Then,∠ADC=∠DCE
∠DCE=65° (32+33)
:) Hope this helps!!!
Answered by
13
let x = 11a and y = 19a
In right ∆BAD, we have A = 90°
also, x+y+A = 180° (angle sum property)
x+y+90°=180°
x+y=90°
11a+19a=90°
30a=90°
a=3°
x=11a = 11×3=33, y=19a = 19×3=57°
since AD||BC and BD cuts them,
DBC = x = 33°
DCE = 32° + DBC
DCE = 32° + 33° = 65°
DCE = 65°
I hope you got your answer
In right ∆BAD, we have A = 90°
also, x+y+A = 180° (angle sum property)
x+y+90°=180°
x+y=90°
11a+19a=90°
30a=90°
a=3°
x=11a = 11×3=33, y=19a = 19×3=57°
since AD||BC and BD cuts them,
DBC = x = 33°
DCE = 32° + DBC
DCE = 32° + 33° = 65°
DCE = 65°
I hope you got your answer
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