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Question

In figure, if $$\angle OAB = 40^{0}$$, then $$\angle ACB$$ equals

1877734_eb592d4439594be1b13d4256b266ff1b.png


A
500
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B
400
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C
600
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D
700
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Solution

The correct option is A $$50^{0}$$

$$\because OA = OB$$ (radius of circle)

$$\therefore \angle OAB = \angle OBA$$ (angles opposite to equal sites)

$$\angle OBA = 40^{0}$$

In right angled $$\Delta OAB$$

$$\angle OAB + \angle OBA + \angle AOB = 180^{0}$$

$$40^{0} + 40^{0} + \angle AOB = 180^{0}$$

$$\angle AOB = 180^{0} – 80^{0}$$

$$\angle AOB = 100^{0}$$

We know that

$$\angle ACB = \dfrac{1}{2}\angle AOB$$

$$=\dfrac{1}{2}\times 100^{0} = 50^{0}$$

Thus, (A) is correct.



Mathematics

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