In the figure, AB || CD and BC || ED. Find the value of x.

1.	In the figure, AB \|\| CD and BC \|\| ED. Find the value of x.
Answer» From the given figure we know that AB and CD are parallel line and BC is a transversal We know that ∠BCD and ∠ABC are alternate angles So we can write it as ∠BCD + ∠ABC = x^o We also know that BC \|\| ED and CD is a transversal From the figure we know that ∠BCD and ∠EDC form a linear pair of angles So it can be written as ∠BCD + ∠EDC = 180^o By substituting the values we get ∠BCD + 75^o = 180^o On further calculation we get ∠BCD = 180^o – 75^o By subtraction ∠BCD = 105^o From the figure we know that ∠BCD and ∠ABC are vertically opposite angles So we get ∠BCD = ∠ABC = x = 105^o ∠ABC = x = 105^o Therefore, the value of x is 105^o.

Answer»

From the given figure we know that

AB and CD are parallel line and BC is a transversal

We know that ∠BCD and ∠ABC are alternate angles

So we can write it as

∠BCD + ∠ABC = x^o

We also know that BC || ED and CD is a transversal

From the figure we know that ∠BCD and ∠EDC form a linear pair of angles

So it can be written as

∠BCD + ∠EDC = 180^o

By substituting the values we get

∠BCD + 75^o = 180^o

On further calculation we get

∠BCD = 180^o – 75^o

By subtraction

∠BCD = 105^o

From the figure we know that ∠BCD and ∠ABC are vertically opposite angles

So we get

∠BCD = ∠ABC = x = 105^o

∠ABC = x = 105^o

Therefore, the value of x is 105^o.

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