Question

In rhombus ABCD,angle AOB=(7z+6), and angle DAO=(5z+1). find the angle CDO and angle DCB

Answer 1

Hence angle CDO=121° and angle DCB=142°

Step-by-step explanation:

Given:

In rhombus ABCD we have ∠AOB=(7z+6), and ∠DAO=(5z+1).  

∠CDO and ∠DCB=?

Diagonals of a rhombus cuts at right angles

So

∠AOB=90°

(7z+6)°=90°

7z=90°-6°

7z=84°

$z=\frac{84}{7}$

z=14°

∠DAO=5z+1

          =5(14°)+1

          =70°+1

∠DAO=71°

Diagonals of a rhombus bisects its interior angles

So if ∠DAO=71°=∠BAO

This way ∠A=71°+71°=142°

All angles are equal is a rhombus

Therefore ∠A=∠B=∠C=∠D=142°

So ∠DCB=142°

Also,

$\angle CDO = \angle \frac{D}{2}$

$\angle CDO = \angle \frac{142\°}{2}$   [Diagonals bisects interior angles]

∠CDO=121°          

Answer 2

Step-by-step explanation:

Hence angle CDO=121° and angle DCB=142°

Step-by-step explanation:

Given:

In rhombus ABCD we have ∠AOB=(7z+6), and ∠DAO=(5z+1).

∠CDO and ∠DCB=?

Diagonals of a rhombus cuts at right angles

So

∠AOB=90°

(7z+6)°=90°

7z=90°-6°

7z=84°

z=\frac{84}{7}z=

7

84

z=14°

∠DAO=5z+1

=5(14°)+1

=70°+1

∠DAO=71°

Diagonals of a rhombus bisects its interior angles

So if ∠DAO=71°=∠BAO

This way ∠A=71°+71°=142°

All angles are equal is a rhombus

Therefore ∠A=∠B=∠C=∠D=142°

So ∠DCB=142°

Also,

\angle CDO = \angle \frac{D}{2}∠CDO=∠

2

D

\angle CDO = \angle \frac{142\°}{2}∠CDO=∠

2

142\°

[Diagonals bisects interior angles]

∠CDO=121°