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The tangent at a point A of a circle with center O intersects the diameter PQ of the circle (when extended) at the point B. If $∠ B A Q = 105 °$ , find $∠ A P Q$ .

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∠ A P Q = 105 °

∠ A P Q = 85 °

∠ A P Q = 75 °

∠ A P Q = 45 °

answer is A.

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Detailed Solution

Given that the tangent at a point A of a circle with center O intersects the diameter PQ of the circle (when extended) at the point B.
We have to find

∠ A P Q .

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 °, ∠ P A Q = 90 °

From figure

∠ B A Q = ∠ B A P + ∠ P A Q

Therefore

∠ B A P = ∠ B A Q − ∠ P A Q ⇒ ∠ B A P = 105 ° − 90 ° ⇒ ∠ B A P = 15 °

As the angle made by the intersection of the radial line and tangent at the circle is

90 °, ∠ B A O = 90 °

From the figure,

∠ P A O = ∠ B A O − ∠ B A P ⇒ ∠ P A O = 90 ° − 15 ° ⇒ ∠ P A O = 75 °

As OA and OP are radii, they are equal. As angles opposite to equal sides are equal,

∠ A P O = ∠ P A O

. Therefore,

∠ A P O = 75 °

. Hence,

∠ A P Q = 75 °

The value is,

∠ A P Q = 75 °

Therefore, the correct option is 1.

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