Answer:
The size of the angle is 60°.
Step-by-step explanation:
Let name the regular hexagon as ABCDEF and the Rhombus as ABOF.
Let angle AFO and angle ABO be y (opposite angles of a rhombus are equal)
To Find :
[tex]\angle OBC = x = ?[/tex]
Solution:
As we know polygon ABCDEF is a regular hexagon then we have
[tex]\textrm{some of all the interior angles} = 720\\ \angle A + \angle B + \angle C + \angle D + \angle E + \angle F = 720\\[/tex]
As it is a regular hexagon all angles are equal
∴ [tex]\angle A + \angle A + \angle A + \angle A + \angle A + \angle A = 720\\[/tex]
[tex]6\angle A = 720\\\angle A = 120\\[/tex]
i.e [tex]\angle B = 120\\\angle F = 120[/tex]
Now quadrilateral ABOF is a rhombus given.
Therefore. Opposite angles of rhombus are equal.
∴ [tex]\angle A = \angle BOF = 120\\and\\\angle AFO = \angle ABO = y (say)[/tex]
Now in a quadrilateral sum of all the angles is 360°
∴ [tex]\angle A + \angle BOF + \angle AFO + \angle ABO = 360\\120 + 120 + y + y = 360\\2y = 120\\y = 60[/tex]
But
[tex]\angle B = \angle ABF + \angle OBC\\120 = y + x\\120 = 60 + x\\x = 60[/tex]
Therefore the size of angle x is 60°
