In the figure, tangent XZ touches the circle with centre . . .

Question : In the figure, tangent XZ touches the circle with centre O at Y. Diamater BA when produced meets XZ at X. If ∠BXY=b and ∠AYX=a

i) Prove that b+2a=90°

ii) If length AY=radius, 3cm of the circle find a,b and the length AX.

Doubt by Zoha

Solution by Saumya S.

Solution : 

i)
∠AYX = a (Given)
∠BXY = b (Given)

∠OYX = 90° [Tangent at any point of the circle is perpendicular to the radius]

In ΔOYX

b+90°+∠XOY=180° (ASP)
∠XOY=180°-90°-b
∠XOY=90°-b

In ΔAOY
OA=OY (Radii of the same circle)
∠OAY=∠OYA (Angle opposite to equal sides of a triangle are equal)

∠OYA=90°-a
∵∠OAY=∠OYA 

∴ ∠OAY=∠OYA = 90°-a

Now, 
In ΔOAY
90°-a+90°-b+90°-a = 180°
270°-2a-b=180°
270°-180°=b+2a
90°=b+2a
b+2a=90°
Hence Proved.

ii) OA=OY=AY=3cm 

∠AOY=∠OAY=∠OYA = 60°
(ΔAOY is an equilateral Triangle)
∠AYX=∠OYX-∠OYA
a=90°-60°

a=30°

Now, 
In ΔAXY
∠AXY+∠AYX=∠OAY [Exterior Angle Property]
b+a=60°
b+30°=60°
b=60°-30°
b=30°

Now, 
a=b=30°

AX=AY [Sides opposite to equal angles of a
triangle are equal]
∵ AY= 3 cm
∴ AX= 3 cm