1) The diagonals of a rectangle ABCD intersect at O .If angles BOC = 68 degree , find angle ODA.

2) ABCD is a rhombus in which the altitude from D to side AB bisects AB . Find the angles of the rhombus.

( here what is altitude explain in 2 ques. and defination of altitude.)

answer it fast plzzz

1)

Given : ABCD is a rectangle and Diagonals AC and BD intersect of O and ∠BOC = 68°

⇒ ∠AOD = ∠BOC = 68° (Vertically opposite angles)

Now in ΔAOD

∠ODA = ∠OAD           (OA = OD as diagonals of a rectangle are equal and bisect each other)

and ∠ODA + ∠OAD + ∠AOD = 180°            (Angle sum property)

⇒ ∠ODA + ∠ODA = 180° – ∠AOD = 180° – 68° = 112°

⇒ 2 ∠ODA

2)

Given : ABCD is a rhombus.

Let DX is the altitude from D to AB

Then AX = BX           ( DX bisects AB)

Now in ΔAXD and ΔBXD

AX = BX

∠AXD = ∠BXD = 90°          (DX is altitude)

DX = DX                (Common)

Thus ΔAXD ΔBXD (by RHS congruency criterion) 

⇒ ∠DAX = ∠DBX

⇒ ∠DAB = ∠DBA         ...      (1)

but diagonal of a rhombus bisects the angles

⇒ ∠CBA = 2∠DBA         ...      (2)

from (1) and (2) we get

∠CBA = 2∠DAB

we know that adjacent angles of a rhombus are supplementary

⇒ ∠DAB + ∠CBA = 180°

⇒ ∠DAB + 2∠DBA = 180°

⇒ 3∠DAB = 180°

and ∠CBA = 2 × 60° = 120°

 also the opposite angles of rhombus are equal

⇒ ∠BCD = ∠DAB = 60°

and ∠ADC = ∠CBA = 120°

Hence

∠A = 60°, ∠B = 120°, ∠C = 60° and ∠D = 120°