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In the figure, AB is diameter of a circle with center $O$ and $QC$ is a tangent to the circle at $C$ . If $∠ CAB =$ $30 °,$ find $∠ CQA$ and $∠ CBA$

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60 o

30 o

50 o

30 o

30 o

30 o

90 o

30 o

answer is C.

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Detailed Solution

Given that AB is diameter of a circle with center

O

and

QC

is a tangent to the circle at

C

.
We have to find

∠ CQA

and

∠ CBA .

The inscribed angle formed by connecting a line from each end of the diameter to any point on the semicircle is equal to

90 o

As the inscribed angle formed by connecting a line from each end of the diameter to any point on the semicircle is equal to,

90 °, ∠ A C B = 90 °

From the figure,

∠ A B C + ∠ B C A + ∠ C A B = 180 ° ⇒ ∠ A B C + 90 ° + 30 ° = 180 ° ⇒ ∠ A B C = 180 ° − 120 ° ⇒ ∠ A B C = 60 °

Therefore,

∠ C B Q = 180 ° − ∠ C B A ⇒ ∠ C B Q = 180 ° − 60 ° ⇒ ∠ C B Q = 120 °

Here, as alternate segments are equal,

∠ C A B = ∠ B C Q

From the figure,

∠ C B Q + ∠ B C Q + ∠ C Q B = 180 ° ⇒ ∠ C Q B + 120 ° + 30 ° = 180 ° ⇒ ∠ C Q B = 180 ° − 150 ° ⇒ ∠ C Q B = 30 °

The values are

∠ C B A = 60 °

and

∠ C Q A = 30 ° .

Therefore, the correct option is 3.

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