O is the circumcentre of triangle ABC. If ABC = 72º. angle ACB = 68 degrees then angle BOC is
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Answer:
(a) Given that,
Arc AB subtends ∠APB at the center and ∠ACB at the remaining part of the circle.
∴∠ACB=
2
1
∠APB=
2
1
×130
o
=65
o
But ∠ACB + ∠BCD = 180
o
(Linear Pair)
⇒65
o
+∠BCD=180
o
⇒∠BCD=180
o
−65
o
=115
o
Major arc BD subtends reflex ∠BQD at the centre and ∠BCD at the remaining part of the circle
∴ reflex ∠BQD=2∠BCD
=2×115
o
=230
0
But reflex ∠BQD+x=360
o
(Angles at a point)
230
o
+x=360
o
⇒x=360
o
−230
o
=130
o
(b) Join OC
In ∆ABC,
AC=BC
⇒∠A=∠B
But ∠A+∠B+∠C=180
o
⇒∠A+∠A+56
0
=180
0
⇒2∠A=180
o
−56
0
=124
o
⇒∠A=
2
1
×124=62
o
⇒∠CAB=62
0
Now, OC is the radius of the circle, and OC bisects ∠ACB
⇒∠OCA=
2
1
∠ACB=
2
1
×56
o
=28
o
Now in △OAC
OA=OC (radii of the same Circle)
⇒∠OAC=∠OCA=28
o
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