Question

prove that angle AOX AND BOY =90

Answer 1

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°

Ab is a line
So <AOB = 180

Angle AOB = angle POA +Angle AOC+COB+OCQ

Angle AOB=<POA+<OBQ+90° (angle aoc +angle cob = 90° )

180 =<POA+<OBQ+90

180-90= <POA+OBQ

<POA+<OBQ = 90°

Hence proved

Answer 2

Step-by-step explanation:

Relax ∠ABC=180−∠CBN

Then angle $$AOC=360- \angle CBA-\angle OAB-\angleO CB$$

Then ∠AOC=360−115−115 (Gven angle CBN =65)

So ∠AOC=130

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