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Question

# Consider the figure where ∠AOB=90∘ and ∠ABC=30∘. Then, ∠CAO = ___.

A
45
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B
50
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C
75
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D
60
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Solution

## The correct option is D 60∘Given that ∠AOB=90∘ and ∠ABC=30∘. We know that the angle subtended by an arc of a circle at the centre is twice the angle subtended by it at any remaining part of the circle. ∴ ∠AOB=2 ∠ACB i.e., 90∘=2∠ACB ⇒ ∠ACB=45∘ Also, AO=OB. ⟹∠ABO=∠BAO [angles opposite to equal sides are equal] .....(i) In ΔOAB, ∠OAB+∠ABO+∠BOA=180∘. [angle sum property of a triangle] ⟹∠OAB+∠OAB+90∘=180∘ [from Eq. (i)] ⇒ 2∠OAB=180∘−90∘ ⇒ ∠OAB=90∘2=45∘ Using angle sum property in ΔACB, we have ∠ACB+∠CBA+∠CAB=180∘. ∴ 45∘+30∘+∠CAB=180∘ ⇒ ∠CAB=180∘−75∘=105∘ But, ∠CAO=∠CAB−∠OAB =105∘−45∘=60∘. i.e., ∠CAO=60∘

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