In the figure, ABCD is a rhombus; DB and AC are the diagonals of the rhombus intersecting at O. If ∠OAB = 35°, then find the value of ∠OCD + ∠OBA
Answers
Answer:
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Step-by-step explanation:
Solution
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:
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Step-by-step explanation:
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