Question

in a rectangle abcd, the diagonals intersect at o. if angle oab=30, find angles a)acb b) abo c)cod d)boc

Answer 1

Answer:

We know that opposite sides of rectangle is equal or parallel.

a) ad is parallel to bc -

ac is transversal line .

then , angle ACB = 30°

b). in triangle abo,

angle oab = 30°

and angle aob = 90°

hence ,

oab + aob + abo =180° ( angle sum property)

• 90° + 30° + abo = 180°
• 120° + abo = 180°
• abo = 180 - 120 = 60 °
• abo = 60°

c). cod = 90 ° ( because angle meet at 90° )

d). boc = 90° also..

l hope it's helpful for all of you

if yes plz mark it

Answer 2

In △ABC,

⇒ ∠CAB+∠ABC+∠ACB=180⁰.

⇒ 30⁰ +90⁰+∠ACB=180⁰.

⇒ 120⁰ +∠ACB=180⁰.

∴ ∠ACB=60⁰

We know that, diagonals of rectangle are equal and bisect each other equally.

∴ AO=OC=BO=OD

In △ABO,

⇒ AO=BO

⇒ ∠OAB=∠ABO [ Angle opposite to equal side are also equal ]

⇒ ∠OAB=∠ABO=30⁰

⇒ ∠OAB+∠ABO+∠BOA=180 ⁰

⇒ 30⁰+30⁰+∠BOA=180⁰.

⇒ ∠BOA=120⁰.

⇒ ∠BOA=∠COD [ Vertically opposite angle ]

∴ ∠COD=120⁰

⇒ ∠COD+∠BOC=180⁰ [ Linear pair ]

⇒ 120⁰+∠BOC=180⁰

∴ ∠BOC=60⁰.

⇒∠ACB=60⁰,∠ABO=300⁰,∠COD=120⁰ and ∠BOC=60⁰.