In the figure, O is the centre and AD is the diameter of the circle. B and C are two point
BAC
on it.
DOB - 50
whether
B
А A
50°
Using a
D
this wo
500
C С
ZAOB =
[ZAOB and BOD are linear pairs]
Write the relation between ZOAB and ZOBA
ZOBA=
ZOAB
[DOB is the outer angle at the corner of the triangle AOB]
Write the suitable name of triangle AOB
Like this, find the ZOAC.
Measure of ZBOC
DOB+
Saino Saltel
.+50°- 100
Answers
Answered by
1
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
.
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