In the figure o is the centre and AD is the diameter of the circle B andC are two points
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In the given figure, AD is a diameter, O is the centre of the circle, AD is parallel to BC and ∠CBD=32
o
Find:
(i) ∠OBD
(ii) ∠AOB
(iii) ∠BED
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Solution
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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
.