Hey Everyone, I really need help on this asap, thank you so much :)

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°