$From the given figure, find ∠ A O B (in degrees ) .$

140

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Solution

The correct option is A 140
We know that, tangents drawn at any point on the circle is perpendicular to the radius through the point of contact.

$∴ O A ⊥ P A and O B ⊥ P B \Rightarrow ∠ O A P = ∠ O B P = 90^{\circ}$

$Also, the sum of all four angles of quadrilateral is 360^{\circ} .$

$∴ ∠ O A P + ∠ A P B + ∠ O B P + ∠ A O B = 360^{\circ} \Rightarrow 90^{\circ} + 40^{\circ} + 90^{\circ} + ∠ A O B = 360^{\circ} \Rightarrow ∠ A O B = 360^{\circ} - 90^{\circ} - 40^{\circ} - 90^{\circ} ∴ ∠ A O B = 140^{\circ}$

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From the given figure, find∠AOB(in degrees ). 140

$From the given figure, find ∠ A O B (in degrees ) .$

140