Question

the given figure, ABCD is a parallelogram in which angle ∠CAD= 40° , ∠BAC=30° and ∠COD=65° 21. Find the value of ∠AOB ​

Answer 1

Answer:

(i)

∠COD=65

o

[ Given ]

⇒ ∠COD=∠AOB [ Vertically opposite angles ]

∴ ∠AOB=65

o

.

In △AOB,

⇒ ∠BAC+∠AOB+∠ABO=180

o

[ Sum of angles of a triangle is 180

o

. ]

⇒ 35

o

+65

o

+∠ABO=180

o

.

⇒ 100

o

+∠ABO=180

o

⇒ ∠ABO=80

o

∴ ∠ABD=80

o

(ii)

AB∥CD abd BD is transversal.

∴ ∠ABD=∠BDC [ Alternate angles ]

∴ ∠BDC=80

o

(iii)

⇒ ∠AOB+∠BOC=180

o

[ Linear pair ]

⇒ 65

o

+∠BOC=180

o

∴ ∠BOC=115

o

⇒ ∠DAO=∠OCB [ Alternate angles ]

∴ ∠OCB=40

o

In △OCB

⇒ ∠BOC+∠OCB+∠CBO=180

o

.

⇒ 115

o

+40

o

+∠CBO=180

o

⇒ 155

o

+∠CBD=180

o

⇒ ∠CBD=25

o

.

Answer 2

Step-by-step explanation:

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