Question

In the figure tangent xy touches the circle with centre o at y. diameter BA when produced meets XZ at X .IF angle bxy=b and angle AYX=a . prove that 2a+b=90

Answer 1

Answer:  OY is perpendicular to xy (Radius is perp. to tangent)

<OYX=90

<AYX +<OYA= 90

a+<oya=90

<oya=90-a -(i)

IN tri.OAY

AO=OY (radii of circle)

<oay=90-a

<xay=180-<oay  (linear pair)

<xay=90+a

Angle sum in TRiangleAXY

2a+b=90

Step-by-step explanation:

Answer 2

Answer:

Since,  the radius from the center of the circle to the point of contact of tangency is perpendicular to the tangent line.

         ∠OYX   =  90°

∠AYX + ∠OYA = 90°

     a   +  ∠OYA = 90°

subtract a from both the sides in above in above expression,

            ∠OYA = 90° - a    

In Δ AOY

AO = OY (radii of circle)

∠OAY = ∠OYA  ( angles opposite to equal sides are equal)

∠OAY = 90°- a

∠XAY +∠OAY = 180° (linear pair)

             ∠XAY  = 180° -∠OAY

             ∠XAY  = 180° - (90°- a)

             ∠XAY  = 90°+ a

In ΔAYX

By Angle sum property

∠XAY +∠AXY +∠AYX = 180°

90°+ a + b + a =  180°

2a + b +  90°=  180°

subtract 90° from both the sides in above expression

2a + b = 90°

Hence proved