In the given figure, AB || CD, ∠BAE = 65° and ∠OEC = 20°. Find

1.	In the given figure, AB \|\| CD, ∠BAE = 65° and ∠OEC = 20°. Find ∠ECO.
Answer» We know that AB \|\| CD and AE is a transversal From the figure we know that ∠BAE and ∠DOE are corresponding angles So we get ∠BAE = ∠DOE = 65^o From the figure we know that ∠DOE and ∠COE form a linear pair of angles So we get ∠DOE + ∠COE = 180^o By substituting the values 65^o + ∠COE = 180^o On further calculation ∠COE = 180^o – 65^o By subtraction ∠COE = 115^o We know that the sum of all the angles in triangle OCE is 180^o. ∠OEC + ∠ECO + ∠COE = 180^o By substituting the values we get 20^o + ∠ECO + 115^o = 180^o On further calculation ∠ECO = 180^o – 20^o – 115^o By subtraction ∠ECO = 45^o Therefore, ∠ECO = 45^o

In the given figure, AB || CD, ∠BAE = 65° and ∠OEC = 20°. Find ∠ECO.

Answer»

We know that AB || CD and AE is a transversal

From the figure we know that ∠BAE and ∠DOE are corresponding angles

So we get

∠BAE = ∠DOE = 65^o

From the figure we know that ∠DOE and ∠COE form a linear pair of angles

So we get

∠DOE + ∠COE = 180^o

By substituting the values

65^o + ∠COE = 180^o

On further calculation

∠COE = 180^o – 65^o

By subtraction

∠COE = 115^o

We know that the sum of all the angles in triangle OCE is 180^o.

∠OEC + ∠ECO + ∠COE = 180^o

By substituting the values we get

20^o + ∠ECO + 115^o = 180^o

On further calculation

∠ECO = 180^o – 20^o – 115^o

By subtraction

∠ECO = 45^o

Therefore, ∠ECO = 45^o