from an external point P two tangents PT and PS are drawn to a circle with centre O and radius r . If OP=2r. show that angle OTS =angle OST=30.
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AP is the tangent to the circle.
∴ OA ⊥ AP (Radius is perpendicular to the tangent at the point of contact)
∠ OAP = 90º
In Δ OAP,
sin ∠OPA= OA/OP=R/2R =1/2
∠OPA=30
In Δ ABP,
AP=BP
∠PAB=∠PBA
so, 60+∠PAB+∠PBA=180
60+2∠PAB=180
∠PAB=180-60/2
∠PAB=60
But
as ∠ OAP=OBP=90
OAP =OBP
so,
60+x=90
x=30
therefore,
∠OTS=OST=30
Thus proved.
∴ OA ⊥ AP (Radius is perpendicular to the tangent at the point of contact)
∠ OAP = 90º
In Δ OAP,
sin ∠OPA= OA/OP=R/2R =1/2
∠OPA=30
In Δ ABP,
AP=BP
∠PAB=∠PBA
so, 60+∠PAB+∠PBA=180
60+2∠PAB=180
∠PAB=180-60/2
∠PAB=60
But
as ∠ OAP=OBP=90
OAP =OBP
so,
60+x=90
x=30
therefore,
∠OTS=OST=30
Thus proved.
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