Question

The diagonals of a rectangle abcd intersect at o if angle aob =114 deegree, find angle acd and angle adb

Answer 1

In Δ AOB ,

AO = OB [ property of a rectangle ]

∠OAB = ∠OBA [ sides of an isosceles Δ ]

==> ∠OAB + ∠OBA + ∠AOB = 180° [ ∠ sum property ]

==> ∠OAB + ∠OAB + 114° = 180°

==> 2 ∠OAB + 114° = 180°[ equal ∠s ]

==> 2∠OAB = 180° - 114°

==> 2∠OAB = 66°

==> ∠OAB = 66°/2

==> ∠OAB = 33°

∠OAB = ∠OBA = 33° [ equal ∠s ]

Now :

==> ∠ACD = ∠OAB [ alternate ∠s ]

==> ∠ACD = 33°

Also :

∠ADB = ∠DBC [ alternate ∠s ]

We know ∠ABC = 90° [ ∠s of rectangle ]

==> ∠ABO + ∠DBC = 90°

==> ∠DBC = 90° - ∠ABO

==> ∠DBC = 90° - 33°

==> ∠DBC = 57°

Hence ∠ADB = ∠DBC = 57° [ stated above]

ANSWERS :

∠ACD = 33°

∠DBC = 57°

Hope it helps

_____________________________________________________________________

Answer 2

Hope this will help u... :-)