In the figure given below, ABCD is a rhombus. Find the value of x and

1.	In the figure given below, ABCD is a rhombus. Find the value of x and y.
Answer» From the figure we know that ∠ A = ∠ C = 62^o Consider △ BCD We get BC = DC It can be written as ∠ CDB = ∠ DBC = y^o Using the sum property of triangle ∠ BDC + ∠ DBC + ∠ BCD = 180^o By substituting the values y + y + 62^o = 180^o On further calculation 2y = 180^o – 62^o By subtraction 2y = 118^o By division y = 59^o We know that the diagonals of a rhombus are perpendicular to each other Consider △ COD as a right angle triangle ∠ DOC = 90^o ∠ ODC = y = 59^o It can be written as ∠ DCO + ∠ ODC = 90^o To find ∠ DCO ∠ DCO = 90^o – ∠ ODC By substituting the values ∠ DCO = 90^o – 59^o ∠ DCO = x = 31^o Therefore, x = 31^o and y = 59^o.

In the figure given below, ABCD is a rhombus. Find the value of x and y.

Answer»

From the figure we know that

∠ A = ∠ C = 62^o

Consider △ BCD

We get

BC = DC

It can be written as

∠ CDB = ∠ DBC = y^o

Using the sum property of triangle

∠ BDC + ∠ DBC + ∠ BCD = 180^o

By substituting the values

y + y + 62^o = 180^o

On further calculation

2y = 180^o – 62^o

By subtraction

2y = 118^o

By division

y = 59^o

We know that the diagonals of a rhombus are perpendicular to each other

Consider △ COD as a right angle triangle

∠ DOC = 90^o

∠ ODC = y = 59^o

It can be written as

∠ DCO + ∠ ODC = 90^o

To find ∠ DCO

∠ DCO = 90^o – ∠ ODC

By substituting the values

∠ DCO = 90^o – 59^o

∠ DCO = x = 31^o

Therefore, x = 31^o and y = 59^o.

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