Question

in the figure PA and PB are tangents to a circle with centre O if angle APB=50 find the angle AOB

Answer 1

Since OA is perpendicular to PA and also, OB is perpendicular to PB

∠APB + ∠AOB = 180°

50°+ ∠AOB = 180°

∠AOB = 180° –  50° = 130°

In △AOB,

OA = OB = radii of same circle

∠OAB = ∠OBA = x ( say )

Again, ∠OAB + ∠OBA + ∠AOB = 180°

x +x + 130° = 180°

2x = 180° –  130° = 50°

X = 25°

Hence, ∠OAB =25°

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Answer 2

Since tangents to a circle is perpendicular to the radius .

∴ OA ⊥ AP and OB ⊥ BP.

⇒ ∠OAP = 90° and ∠OBP = 90°

⇒ ∠OAP + ∠OBP  = 90° + 90° = 180°  ........... (1)

In quadrilateral OAPB,

∠OAP + ∠APB + ∠AOB + ∠OBP = 360°

⇒ (∠APB + ∠AOB) + (∠OAP + ∠OBP) = 360°

⇒ ∠APB + ∠AOB + 180° = 360°  [ From (1) ]

⇒ ∠APB + ∠AOB = 180°  ................ (2)

From (1) and (2), the quadrilateral AOBP is cyclic.

TO PROVE,

<A+<B=180

<P+<O=180  [SINCE SUM OF OPPOSITE ANGLES OF A CYCLIC                   QUARDRILATERAL IS 180]

PROOF

WE KNOW THAT <A=<B=90[THE RADIUS IS PERPENDICULAR TO THE TANGENT AT THE POINT OF CONTACT.

SO <A+<B=90+90=180

WE KNOW THAT SUM OF ANGLES OF A QUARDRILATERAL IS 360.THEREFORE

<A+<B+<P+<O=360

=90+90+<P+<O=360

=<P+<O=360-180

=<P+<O=180

HENCE THE PROOF.

<A+<B=180