Question

In the given figure tangent XZ touches the circle with centre O at point Y. The diameter BA is extended up to point X if angle BXY =b and angle AYX=a then prove thatb+2a=90

Answer 1

Solution:

In Δ OYX

OA=OY→→radii of circle

∠OAY=∠OYA=x degree→→If sides in a triangle are equal angle opposite to these sides are equal.

∠OYX=90°→→Line from center of circle to point of contact of tangent makes an angle of 90° at the point of contact of tangent.

x+a=90°------(1)

∠AYX=a

∠BXY =b

OAX lie in a line.

∠OAY +∠YAX=180°→→by linear pair

∠YAX=180°-x

In ΔYAX

a+b+180°-x=180°→→Angle sum property of triangle

a+b=x

a+b=90°-a→→→using 1

a+a+b=90°

2a +b=90°

Hence proved.

Answer 2

Step-by-step explanation:

For explanation see the 2 attachments.....

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