Question

solve the above questions in the attachment
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Answer 1

Const - join O and C

A/q,

In ΔOPA and ΔOCA,

OP = OC (Radii of the same circle)

AP = AC (Tangents from point A)

AO = AO (Common side)

∴ ΔOPA ≅ ΔOCA (SSS congruence criterion)

⇒ ∠POA = ∠COA … (i)

Similarly,

ΔOQB  ≅ ΔOCB

∠QOB = ∠COB … (ii)

Since POQ is a diameter of the circle, it is a straight line.

∴ ∠POA + ∠COA + ∠COB + ∠QOB = 180 º

From equations (i) and (ii),

2∠COA + 2∠COB = 180º

⇒ ∠COA + ∠COB = 90º

⇒ ∠AOB = 90° proved