The following figure ABCD is a parallelogram. Find the measures of
(i) ∠ABD (ii) ∠BDC (iii) ∠ABC (iv) ∠CBD. plz don't spam
Attachments:
Answers
Answered by
0
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
Similar questions