In the given figure, AB and CD are two parallel tangents to a circle with center O. ST is tangent segment between the two parallel tangents touching the circle at Q. Show that angle SOT =90°.
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Answer:
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Step-by-step explanation:
Join OP and OQ.
In OPS and OQS:
OP=OQ……Radius of circle
OPS=OQS=900 (Tangent is to the radius through the point of contact)
OS=OS….common
Therefore, by Right angle-Hypotenuse-Side criterion of congruence, we have
OPS OQS (RHS)
The corresponding parts of the congruent triangles are congruent.
POS = QOS……cpct,
OSP = OSQ ……..cpct
In quadrilateralPSQO:
POQ+PSQ=1800 as the total sum of interior angles in a quadrilateral=3600
POQ=PSQ=900
So OSP = OSQ = 45oAs PSQ=900
STC=900 sum of two co interior angles is 180 o.
SOT is an isosceles triangle having two sides:
OS=OT
Similarly, QOT = 45o
SOT = 90o
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