A and B are the ends of the diameter of a circle of centre O TP is a tangent - Maths - Circles

Question

A and B are the ends of the diameter of a circle of centre O TP is
a tangent drawn from external point T to a point on the circle P If
ang PBT = 30o, prove that BA:AT =2:1

Akhil Goyal · Accepted Answer

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Let r be the radius of the circle:Consider △PBT and △APT (note the sequencing of points)We know by the Tanget-secant rule that: (PT)2=AT·BT⇒PTBT=ATPTalso, ∠T is common for both triangles
Hence, △PBT&△APT are similar triangles
Now, ∠PBT=30⇒∠APT=30 (corresponding angles of similar triangles are equal)Also, BO=PO=r⇒∠PBT=∠OPB=30 (since opposite angles of opposite sides are equal)⇒∠TPB=∠BPA+∠APT=90+30=120 (since ∠BP=90, as PT is tangent to the circle) ⇒∠T=(180-120-30)=30Since ∠APT=30 ⇒∠APT=∠T⇒PA=TAbut, also ∠APO=90-30=60and, OP=OA=r ⇒∠PAO=∠APO=60Therefore, △POA is equilateral and thus, PA=r⇒PA=AT=rBA=2rThus, BAAT=21

A and B are the ends of the diameter of a circle of centre O. TP is a tangent drawn from external point T to a point on the circle P. If ang.PBT = 30o, prove that BA:AT =2:1.

A and B are the ends of the diameter of a circle of centre O. TP is a tangent drawn from external point T to a point on the circle P. If ang.PBT = 30^o, prove that BA:AT =2:1.