Question

If PA and PB are two tangents from external point P to a circle with centre O and [tex]angle
[/tex]APB = 35$degree$ , find the angle OAB.

Answer 1

$In\ \Delta OAP, \angle OAP=90^0\ (PA\ is\ tangent)\\similarly,\ In\ \Delta OBP, \angle OBP=90^0\\given\ that\ \angle APB=35^0\\ \angle AOB=360-90-90-35=145^0\\ in\ \Delta OAB,\ AO=BO=radius\\so\ \Delta OAB\ is\ an\ isosceles\ triangle\\thus\ \angle OAB=\angle OBA\\ \\in\ \Delta OAB;\\ \angle OAB+\angle OBA+\angle AOB=180\\\angle OAB+\angle OAB+145=180\\2 \angle OAB=180-145=35\\ \angle OAB= \frac{35}{2}\\\angle OAB=\boxed{17.5^0}$

Answer 2

Answer:

∠OAB = 17.5°

Step-by-step explanation:

For better understanding of the solution see the attached figure :

Given : ∠APB = 35°

To find : ∠OAB

Solution : ∠PAO = ∠PBO= 90 ( Tangent makes right angles with the radius of the circle)

Now, in quadrilateral sum of all angles is 360

⇒ ∠APB + ∠PAO + ∠PBO + ∠AOB = 360

⇒ 35 + 90 + 90 + ∠AOB = 360

⇒ ∠AOB = 145°

Now, since OA = OB ( Radius of the same circle)

⇒ ∠OAB = ∠OBA ( Angles opposite to equal sides are equal)

Now, in ΔAOB using angle sum property of a triangle

∠AOB + ∠OBA + ∠OAB = 180

⇒ 145 + ∠OAB + ∠OAB = 180

⇒ 2∠OAB = 35

⇒ ∠OAB = 17.5°