Question

Hii...Guysss plzzzz answer it ....
Plzz answer Q20 plzzzz.....

Answer 1

PA and PB are the tangents to the circle.

∴ OA ⊥ PA

⇒ ∠OAP = 90°

In ΔOPA,

 sin ∠OPA = OA OP  =  r/2r   [Given OP is the diameter of the circle]

⇒ sin ∠OPA = 1/2 =  sin  30 ⁰

⇒ ∠OPA = 30°

Similarly, it can be proved that ∠OPB = 30°.

Now, ∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°

In ΔPAB,

PA = PB         [lengths of tangents drawn from an external point to a circle are equal]

⇒∠PAB = ∠PBA ............(1)   [Equal sides have equal angles opposite to them]

∠PAB + ∠PBA + ∠APB = 180°    [Angle sum property]

⇒∠PAB + ∠PAB = 180° – 60° = 120°  [Using (1)]

⇒2∠PAB = 120°

⇒∠PAB = 60°    .............(2)

From (1) and (2)

∠PAB = ∠PBA = ∠APB = 60°

∴ ΔPAB is an equilateral triangle.

Answer 2

HELLO DEAR,

JOIN. A. TO. B

WE HAVE

op = diameter

or

OP+OQ=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.

I HOPE ITS HELP YOU DEAR,
THANKS