solve the question
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Answers
Answer:
For triangle ABC
angle ABC = 50° angle BAC = 30°
so angle ACB = 180 °-(50+30)
[THE TOTAL SUM UP OF THE ANGLE IN ANY TRIANGLE IS 180°]
=180°-80°
=100°
Now angle ECD is =180°-100°=80°
angle CDE = 35° So angle DEC = 180°-(80°+35°) = 180°-115°=65°
so angle x = 180°-<DEC=180°-65°=115°
Answer :
x = 115°
Step-by-step explanation :
In ΔABC :
∠ABC = 50°
∠BAC = 30°
Let ∠ACB = a°
The sum of all the angles in a triangle = 180°
∠ABC + ∠BAC + ∠ACB = 180°
50° + 30° + a° = 180°
80° + a° = 180°
a° = 180° - 80°
a° = 100°
∴ ∠ACB = 100°
______________________
Let ∠ECD = b°
Now, ∠ACB and ∠ECD are on a straight line.
The sum of the angles on a straight line = 180°
∠ACB + ∠ECD = 180°
100° + b° = 180°
b° = 180° - 100°
b° = 80°
∴ ∠ECD = 80°
_______________________
In ΔECD :
∠ECD = 80°
∠ EDC = 35°
Let ∠CED = c°
The sum of all the angles in a triangle = 180°
∠ECD + ∠EDC + ∠CED = 180°
80° + 35° + c° = 180°
115° + c° = 180°
c° = 180° - 115°
c° = 65°
∴ ∠CED = 65°
_______________________
∠CED and ∠DEA are on a straight line.
∠DEA = x° (given)
The sum of the angles on a straight line = 180°
∠CED + ∠DEA = 180°
65° + x° = 180°
x° = 180° - 65°
x° = 115°
∴ ∠DEA = 115°
