In the adjoining figure ABCD is a rectangle with < OAB = 30° find <ABO and <COD
Answers
Step-by-step explanation:
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
.
Wish this helps you please mark me brainliest ☺️
Step-by-step explanation:
<OAB = < ABO = 30 ° OPPOSITE ANGLES
<COD = <AOB
<AOB = 180 - <OAB + < ABO
<AOB = 180 - 30 + 30 = 120
<COD = 120°