Q.

In the figure XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C interesting XY at A and X'Y' at B, what is the measure of ∠AOB.


Ans.

(90°)

Join OC. Since, the tangents drawn to a circle from an external point are equal.

∴ AP = AC

In ∆PAO and ∆AOC, we have:

AO = AO [Common]

OP = OC [Radii of the same circle]

AP = AC

⇒∆PAO≅∆AOC   [SSS Congruency]

∴∠PAO=∠CAO=∠1

∠PAC=2∠1                              ...(i)

Similarly ∠CBQ=2∠2               ...(2)

Again, we know that sum of internal angles on the same side of a transversal is 180°.

∴ ∠PAC+∠CBQ=180°

⇒2∠1+2∠2=180°  [From (1) and (2)]

⇒∠1+∠2=180°2=90°      ....(3)

Also ∠1+∠2+∠AOB=180°  [Sum of angles of a triangle]

⇒90°+∠AOB=180°

⇒∠AOB=180°-90°

⇒∠AOB=90°