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If PA and PB are tangents to the circle with centre O such that ∠APB = 40∘, then ∠OAB is equal to
- a)30∘
- b)40∘
- c)25∘
- d)20∘
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
If PA and PB are tangents to the circle with centre O such that∠AP...
Let ∠OAB = ∠OBA = x [Opposite angles of opposite equal radii] And ∠AOB =180° - 40° = 140°
Now, in triangle AOB,
∠OAB + ∠OBA + ∠AOB = 180°
⇒ x + x +140° = 180°
⇒ 2x = 40°
⇒ x = 20°
∴ ∠OAB = 20°
Now, in triangle AOB,
∠OAB + ∠OBA + ∠AOB = 180°
⇒ x + x +140° = 180°
⇒ 2x = 40°
⇒ x = 20°
∴ ∠OAB = 20°
Most Upvoted Answer
If PA and PB are tangents to the circle with centre O such that∠AP...
Given, PA&PB are two tangent and o is the centre of circle
TO FIND:⟨OAB=?
In ∆OAB,
OA=OB [radius]
⟨OAB=⟨OBP(angle opposite to equal side are equal)
since, OA perpendicular PA [radius perpendicular tangent]
so,⟨OAP=90⁰
similarly,
⟨OBP=90⁰
In Quadrilatera OAPB,
⟨A + ⟨B + ⟨P + ⟨O=360⁰[sum of all interior angle=360⁰]
90⁰+90⁰+40⁰+ ⟨O=360⁰
220⁰+ ⟨O=360⁰
⟨O=140⁰
Again,In ∆OAB
⟨AOB + ⟨OAB + ⟨OBA=180⁰
140⁰+ ⟨OAB+⟨OAB=180⁰
2⟨OAB=180⁰-140⁰
⟨OAB=40⁰/2
Therefore,⟨OAB=20⁰
TO FIND:⟨OAB=?
In ∆OAB,
OA=OB [radius]
⟨OAB=⟨OBP(angle opposite to equal side are equal)
since, OA perpendicular PA [radius perpendicular tangent]
so,⟨OAP=90⁰
similarly,
⟨OBP=90⁰
In Quadrilatera OAPB,
⟨A + ⟨B + ⟨P + ⟨O=360⁰[sum of all interior angle=360⁰]
90⁰+90⁰+40⁰+ ⟨O=360⁰
220⁰+ ⟨O=360⁰
⟨O=140⁰
Again,In ∆OAB
⟨AOB + ⟨OAB + ⟨OBA=180⁰
140⁰+ ⟨OAB+⟨OAB=180⁰
2⟨OAB=180⁰-140⁰
⟨OAB=40⁰/2
Therefore,⟨OAB=20⁰
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Community Answer
If PA and PB are tangents to the circle with centre O such that∠AP...
We know that the sum of opposite angles in a circle is 180 degrees
so, APB+AOB=180
40+AOB=180
AOB=140
then by Angle sum property
140+OAB+OBA=180
so,OAB=20degrees
so, APB+AOB=180
40+AOB=180
AOB=140
then by Angle sum property
140+OAB+OBA=180
so,OAB=20degrees
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If PA and PB are tangents to the circle with centre O such that∠APB = 40∘, then∠OAB is equal toa)30∘b)40∘c)25∘d)20∘Correct answer is option 'D'. Can you explain this answer?
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