Question

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CLASS - X
CH - 10
CIRCLES ​

Answer 1

angle OXY + angle XYO + angle O =180° (sum angle property)

b + 90° (as tangent is perpendicular to radius) + angle XOY = 180°

angle XOY = 180° -90°-x

angle XOY =90° - x

Now, OA = AY (radii of same circle)

So, angle AYO = angle XOY or AOY= 90°-x


In ∆ OAY

angle OAY+ angle AYO+ angle AOY =180° (sum angle property)

angle OAY + angle AYO + angle AYO =180°

angle OAY +2 angle AYO =180°

90-b +2 AYO=180°

2 angle AYO =180 -90 +b

angle AYO = 90+b/2 = angle OYA ---(1)

Now, Angle AYX = a and Angle OYX =90°

angle OYA =90 - a= OAY ---(2)


From 1 and 2

$\frac{90 + b}{2} = 90 - a$



90+b =2(90-a)

90+b=180-2a

b+2a =180-90

b +2a =90

Hence proved

Answer 2

Angle OXY + angle XYO + angle O =180° (sum angle property)

b + 90° + angle XOY = 180°

angle XOY = 180° -90°-x

angle XOY =90° - x

Now, OA =AY

So, angle AYO = angle XOY or AOY= 90°-x

In ∆ OAY

angle OAY + angle AYO + angle AYO =180°

angle OAY +2 angle AYO =180°

90-b +2 AYO=180°

2 angle AYO =180 -90 +b

angle AYO = 90+b/2 = angle OYA ---(1)

Now, Angle AYX = a and Angle OYX =90°

angle OYA =90 - a= OAY ---(2)

From 1 and 2

90+b =2(90-a)

90+b=180-2a

b+2a =180-90

b +2a =90

Hence proved