In the adjoining figure ABCD, diagonals intersect at O. If angle OAB=30* find angle ACB, ABO, COD, BOC
Answers
Answer:
In △ABC,
⇒ ∠CAB+∠ABC+∠ACB=180
o
.
⇒ 30
o
+90
o
+∠ACB=180
o
.
⇒ 120
o
+∠ACB=180
o
.
∴ ∠ACB=60
o
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=30
o
⇒ ∠OAB+∠ABO+∠BOA=180
o
⇒ 30
o
+30
o
+∠BOA=180
o
.
⇒ ∠BOA=120
o
.
⇒ ∠BOA=∠COD [ Vertically opposite angle ]
∴ ∠COD=120
o
⇒ ∠COD+∠BOC=180
o
[ Linear pair ]
⇒ 120
o
+∠BOC=180
o
∴ ∠BOC=60
o
.
⇒ ∠ACB=60
o
,∠ABO=30
o
,∠COD=120
o
and ∠BOC=60
o
.
Answer:
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.