In the given figure below, AD is a diameter. O is the centre of the circle. AD is parallel to
BC and CBD = 32º.
E
D
32°
Find:
() OBD
(ii) AOB
(ii) BED
Answers
Answer:
don't know sorry br9ther
Answer:
As AD∥BC and DB is a transversal.
⇒ ∠ODB=∠DBC [ Alternate angles ]
∴ ∠ODB=32
o
In △OBD
OB=OD [ Radius of a circle ]
∴ △OBD is isosceles triangle.
⇒ ∠OBD=∠ODB= [ Base angles are equal in isosceles triangle ]
∴ ∠OBD=32
o
∠OBD+∠ODB+∠BOD=180
o
[ Sum of angles of triangle is 180
o
. ]
⇒ 32
o
+32
o
+∠BOD=180
o
⇒ 64
o
+∠BOD=180
o
⇒ ∠BOD=116
o
⇒ ∠BOD+∠AOB=180
o
[ Linear pair ]
⇒ 116
o
+∠AOB=180
o
∴ ∠AOB=64
o
In △AOB,AO=OB, hence its an isosceles triangle
⇒ ∠OAB=∠OBA
Now, ∠AOB+∠OAB+∠OBA=180
o
⇒ 64
o
+∠OAB+∠OAB=180
o
⇒ 2∠OAB=116
o
∴ ∠OAB=58
o
⇒ ∠OAB=∠BED [ Angles subtended by the same chord BD ]
∴ ∠BED=58
o
.
(i) ∠OBD=32
o
(ii) ∠AOB=64
o
(iii) ∠BED=58
o
.