In a rectangle ABCD diagonals intersect at O ,if angle OAB=30 .find angle ACB and angle COD
Answers
Step-by-step explanation:
In \triangle ABC,△ABC,
\Rightarrow⇒ \angle CAB+\angle ABC+\angle ACB=180^o.∠CAB+∠ABC+∠ACB=180
o
.
\Rightarrow⇒ 30^o+90^o+\angle ACB=180^o.30
o
+90
o
+∠ACB=180
o
.
\Rightarrow⇒ 120^o+\angle ACB=180^o.120
o
+∠ACB=180
o
.
\therefore∴ \angle ACB=60^o∠ACB=60
o
We know that, diagonals of rectangle are equal and bisect each other equally.
\therefore∴ AO=OC=BO=ODAO=OC=BO=OD
In \triangle ABO△ABO,
\Rightarrow⇒ AO=BOAO=BO
\Rightarrow⇒ \angle OAB=\angle ABO∠OAB=∠ABO [ Angle opposite to equal side are also equal ]
\Rightarrow⇒ \angle OAB=\angle ABO=30^o∠OAB=∠ABO=30
o
\Rightarrow⇒ \angle OAB+\angle ABO+\angle BOA=180^o∠OAB+∠ABO+∠BOA=180
o
\Rightarrow⇒ 30^o+30^o+\angle BOA=180^o.30
o
+30
o
+∠BOA=180
o
.
\Rightarrow⇒ \angle BOA=120^o.∠BOA=120
o
.
\Rightarrow⇒ \angle BOA=\angle COD∠BOA=∠COD [ Vertically opposite angle ]
\therefore∴ \angle COD=120^o∠COD=120
o
\Rightarrow⇒ \angle COD+\angle BOC=180^o∠COD+∠BOC=180
o
[ Linear pair ]
\Rightarrow⇒ 120^o+\angle BOC=180^o120
o
+∠BOC=180
o
\therefore∴ \angle BOC=60^o.∠BOC=60
o
.
\Rightarrow⇒ \angle ACB=60^o,\,\angle ABO=30^o,\,\angle COD=120^o∠ACB=60
o
,∠ABO=30
o
,∠COD=120
o
and \angle BOC=60^o.∠BOC=60
o
.
In △ABC,
⇒ ∠CAB+∠ABC+∠ACB=180⁰.
⇒ 30⁰ +90⁰+∠ACB=180⁰.
⇒ 120⁰ +∠ACB=180⁰.
∴ ∠ACB=60⁰
⇒ ∠OAB=∠ABO=30⁰
⇒ ∠OAB+∠ABO+∠BOA=180 ⁰
⇒ 30⁰+30⁰+∠BOA=180⁰.
⇒ ∠BOA=120⁰.
⇒ ∠BOA=∠COD [ Vertically opposite angle ]
∴ ∠COD=120⁰