From a point p two tangents pt and ps are drawn to a circle with centre o such that angle spt = angle
RishabhBansal:
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Consider ΔOPS and ΔOPT
OS = OT ( radii)
∠OSP = ∠OTP = 90 (tangents are perpendicular to the radii)
SP = ST ( tangents to a circle from the external point are congruence)
ΔOPS ≅ ΔOPT ( By SAS criterion)
The corresponding parts of the corresponding triangles are congruent.
∠OPS = ∠OPT
since ∠SPT = 120° and ∠OPS = ∠OPT
we have ∠OPS = ∠OPT = 60°
∠POS = ∠POT = 30°
Consider In a ΔPOS
sin 30° = PS / OP
1 / 2 = PS / OP
OP = 2PS.
OS = OT ( radii)
∠OSP = ∠OTP = 90 (tangents are perpendicular to the radii)
SP = ST ( tangents to a circle from the external point are congruence)
ΔOPS ≅ ΔOPT ( By SAS criterion)
The corresponding parts of the corresponding triangles are congruent.
∠OPS = ∠OPT
since ∠SPT = 120° and ∠OPS = ∠OPT
we have ∠OPS = ∠OPT = 60°
∠POS = ∠POT = 30°
Consider In a ΔPOS
sin 30° = PS / OP
1 / 2 = PS / OP
OP = 2PS.
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this can also be done through trigonometry which i did
i donot know to do that way but it can be done
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16
hope it helps pls mark me brainliest..
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