(1) x° = 38° (2) x° = 54° (3) x° = 4°
Step-by -step explanation:
(1) Given ABCD is a rhombus.
Let us take a triangle AED.
Diagonals are perpendicular in rhombus – – – (i)
[tex]\angle \mathrm{AED}=90^{\circ}[/tex]
Sum of the adjacent angles in a triangle is 180°.
[tex]\angle \mathrm{EAD}+\angle \mathrm{ADE}+\angle \mathrm{AED}=180^{\circ}[/tex]
⇒ x° + (x + 14)° + 90° = 180°
⇒ 2x° = 76°
⇒ x° = 38°
(2) Given ABCD is a rhombus, [tex]\angle \mathrm{CED}=90^{\circ}[/tex] (by (i))
[tex]\angle \mathrm{CDE}+\angle \mathrm{DCE}+\angle \mathrm{CED}=180^{\circ}[/tex]
⇒ x° + (x – 18)° + 90° = 180°
⇒ 2x° = 108°
⇒ x° = 54°
(3) Given ABCD is a rhombus.
BD bisect the angle [tex]\angle \mathrm{ADC}[/tex].
⇒ 8x° = 32
⇒ x° = 4°
