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Knowledge Expert
· commented
· 1 Months ago
Already solution has been given.
Alternate solution:
O is the centre of the circle.
Given, <POQ = 130º
PT and QT are tangents drawn from external point T to the circle
<OPT = <OQT = 90º [ Radius is perpendicular to the tangent at point of contact]
In quadrilateral OPTQ,
<PTQ + <OPT + <OQT + <POQ = 360º
=> <PTQ + 90º + 90º + 130º = 360º
=> <PTQ = 360º – 310º = 50º
Thus, the angle between the tangents is 50º.
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