Question

the diagonal of a rectangle abcd intersect at o if angle aob=114 find angle acd and angle adb

Answer 1

AOB=114

OA=OB=OC=OF

DIAGONALS OF RECTANGLE ARE EQUAL AND BISECT EACH OTHER

ANGLE OAB=OBA

ANGLE OPPOSITE TO EQUAL SIDES ARE EQUAL

SO... AOB+OAB+OBA=180

114+a+a=180

2a=180-114

2a=66

a=66÷2

a=33

ABO=33

OAB=33

ACD=33

BECAUSE OAB AND ACD ARE EQUAL BECAUSE THEY ARE ALTERNATE INTERIOR ANGLES.....

Answer 2

Hello Mate!

Since, diagonals of rectangle are equal and bisect each other so

AC = BD

1 AC / 2 = 1 BC / 2

OC = OD

Therefore, COD is isosceles triangle

Hence, < OCD = < ODC

< COD = 114° [ Vertically opposite angle ]

< COD + < OCD + < ODC = 180° [ Angle sum pro. ]

Let < OCD = x°

x° + x° = 180° - 114°

2x = 66°

x = 33°

Hence, < ODC = < OCD or < ACD = 33°

Since, AB || CD where BD is transeversal.

< ODC = < ABO or < ABD [ Alternate angle ]

< ABD = 33°

Have great future ahead!