8. In figure, if LAOB = 90° and LABC = 30°, then 2CAB is equal to
question of class 9
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Answer:
In △AOB
OA=OB
∴∠OAB+∠OBA+∠AOB=180
0
=>2∠OAB+90
0
=180
0
=>2∠OAB=180
0
−90
0
=90
0
=>∠OAB=
2
90
=45
0
We know that the angle inscribed at the centre of the circle by an arch is twice the angle subtended at the circumference by the same arch.
∴∠ACB=
2
1
∠AOB
=
2
90
=45
0
Again in △ABC
∠ACB+∠CBA+∠BAC=180
0
=>45
0
+30
0
+∠BAC=180
0
=>∠BAC=180
0
−75
0
=105
0
∴∠CAO=∠BAC−∠OAB
=>∠CAO=105
0
−45
0
=60
0
Step-by-step explanation:
I hope it will help you
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