Question

answer me question number 19 hurry up??

Answer 1

⇄ From figure: ⇆

PA and PB are tangents to the circle.

∴ OA ⊥ AP [Tangent at any point of circle is perpendicular to the radius through point of contact]

∴ ∠OAP = 90°.

In ΔOPA,

⇒ sin θ = OA/OP

⇒ sin θ = r/2r

⇒ sin θ = 1/2

⇒ sin θ = sin 30

⇒ θ = 30

So, ∠OPA = 30°.

Similarly, ∠OPB = 30°{Since, ∠OPA = ∠OPB}

Hence, ∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°.   --- (i)

Now,

In ΔPAB,

PA = PB [Length of tangents from an external point is equal]

⇒ ∠PAB = ∠PBA   ---- (ii)

⇒ ∠PAB + ∠PBA + ∠APB = 180°

⇒ ∠PAB + ∠PAB + 60 = 180 {from (i),(ii)}

⇒ ∠PAB + ∠PAB = 120°

⇒ 2∠PAB = 120

⇒ ∠PAB = 60°    ----- (iii)

From (ii),(iii), we get

⇒ ∠PAB = ∠PBA = ∠APB = 60°

Therefore, ΔABP is Equilateral triangle.

Hope it helps!

Answer 2

Join A to B.

We have

OP= Diameter

Or, OQ + OP= diameter

Or, Radius + PQ = diameter (that is OQ=radius)

Or, PQ = diameter – radius

Or, PQ = radius

Or, OQ = PQ = radius

Thus OP is the hypotenuse of the right angle triangle AOP.

So, in ∆AOP sin of angle P = OP /AO

= 1/2

So, P = 30°

Hence, APB = 60°

Now, As in ∆AOP, AP=AB

So, PAB = PBA =60°

Hence,

∆ABP is equilateral triangle.