Respuesta :

Look into my attachment

in rhombus ABCD, ∠AOB = 90°

Look at ΔAOB
the sum of inner angles = 180°
∠BAO + ∠AOB + ∠ABO = 180°
2x + 1 + 90 + 2x + 5 = 180°
4x + 96 = 180
4x = 84
x = 21
Ver imagen gustanika

Answer:

Option C

[tex]21\°[/tex]

Step-by-step explanation:

we know that

The two diagonals of a rhombus are perpendicular

so

Let

O------> the center of the rhombus

m∠AOB=[tex]90\°[/tex]

Remember that

The sum of the internal angles of a triangle is equal to [tex]180\°[/tex]

therefore

in the triangle AOB

m∠AOB+m∠OAB+m∠OBA=[tex]180\°[/tex]

substitute the values

[tex]90\°+(2x+1)\°+(2x+5)\°=180\°[/tex]

solve for x

[tex](4x+6)\°=90\°[/tex]

[tex]4x=90\°-6\°[/tex]

[tex]x=21\°[/tex]