Q1 of 31 Page 209

Masum has drawn a circle with centre ‘O’ of which AB is a chord. I draw a tangent at the point B which intersect extended AO at the point T. If ∠BAT = 21°, let us write by calculating the value of ∠BTA.

Theory.


⇒ Exterior angle is sum of opposite interior angles


⇒ if 2 sides of triangle are equal then their corresponding angles will also be equal


Solution.



In Δ AOB


AO = OB ∵ (radius of same circle )


∴ ∠OAB = ∠OBA = 21°


In Δ BOT


∠BOT = ∠OAB + ∠OBA ∵ (Exterior angle property)


= 21° + 21° = 42°


∠OBT = 90° ∵ (Radius of circle from point of contact of tangent is 90° )


∠OBT + ∠BOT + ∠BTO = 180° ∵ (Angle sum property)


90° + 42° + ∠BTO = 180°


∠BTO = 180° - 132° = 48°


∠BTO = ∠BTA = 48°


More from this chapter

All 31 →
2

XY is a diameter of a circle. PAQ is a tangent to the circle at the point. ‘A’ lying on the circumference. The perpendicular drawn on the tangent of the circle from X intersects PAQ at Z. Let us prove that XA is a bisector of ∠YXZ.

 3

I drew a circle having PR as a diameter. I draw a tangent at tangent at the point P and a point S is taken on the tangent of the circle in such a way that PR = PS. If RS intersects the circle at the point T. Let us prove that ST = RT = PT.

 4

Two radii OA and OB of a circle with centre O are perpendicular to each other. If two tangents drawn at the point A and B intersect each other at the point T, let us prove that AB = OT and they bisect each other at a right-angle.

 5

Two chords AB and AC of the larger of two concentric circles touch the other circle at points P and Q respectively. Let us prove that, PQ = 1/2 BC.