Question

In figure , CA and CB are tangents
to the circle. Also PA=PB and

Answer 1

Answer:

ohhh same question I was gonna ask this same too xD

Answer 2

$\huge{\dag} \: {\underline{\boxed{\sf{\red{ ANSWER \ :-}}}}}$

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Consider the given figure( attachment)

The lengths of two tangents drawn from an external point to a circle are equal.

Thus, CA=CB

Consider △ABC,

Since,

 CA=CB,△ABC is an isosceles triangle.

Thus, the base angles, ∠CAB=∠CBA

★By angle sum property

we have,

m∠CAB+∠CBA+∠ACB=180∘

⇒m∠CAB+m∠CAB+m∠ACB=180∘

⇒2m∠CAB=40∘=180∘

⇒2m∠CAB=140∘⇒m∠CAB=70∘=m∠CBA

Let O be the centre of the circle and OB be the radius.

A tangent to a circle is perpendicular to the radius through the point of contact.

⇒OB⊥CB and OA⊥CA

⇒m∠OBC=90∘ and m∠OAC=90∘

Consider angles, ∠OBA and ∠OAB

m∠OBA=m∠OBC−m∠CBA=90∘−70∘=20∘

Similarly, m∠OAB=m∠OAC−m∠CAB=90∘−70∘=20∘

Consider △AOB,

Since OA and OB are radii, OA=OB

⇒△AOB is an isosceles triangle,

★By angle sum property

we have,

m∠OBA+m∠OAB+m∠AOB=180∘

⇒20∘+20∘+m∠AOB=180∘⇒AOB=140∘

The angle subtended by an arc of a circle at the centre is twice the measure of the angle subtended by it at any point on the remaining part of the circle.

m∠=140∘⇒m∠APB=70∘

Also, given that PA=PB,△APB is an isosceles triangle.

⇒anglePBA=∠PAB

★By angle sum property

we have,

m∠APB+m∠PBA+m∠PAB=180∘⇒70∘+m∠PBA+m∠PBA=180∘

⇒70∘+2m∠PBA=180∘⇒2m∠PBA=110∘⇒m∠PBA=55∘=m∠PAB

Thus, the angles of △APB are m∠PAB=55∘,m∠PBA=55∘,m∠PBA=55∘ and m∠APB=70∘

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