o is the centre of the circle and TP is the tangent to the centre fro m an external point T. IF angle PBT= 30 degree prove that BA; AT = 2; 1
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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⇒PT/BT=AT/PT also,
∠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, BA/AT=2/1
∠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, BA/AT=2/1
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