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In the figure, PQ is a diameter of the circle and XY is chord equal to the radius of the circle. PX and QY when extended intersect at point E. Prove that PEQ = 60o

Asked by Topperlearning User 4th June 2014, 1:23 PM
Answered by Expert
Answer:

PQ is the diameter of the circle, chord XY = r (radius of circle)

PX and QY extended intersect at a point E.

To prove PEQ = 60o

XY = OX = OY [radii of a circle, Given]

XOY is an equilateral triangle

XQY = 30o [Inscribed angle is half of the central angle]

PXQ = 90o [Angle in a semi circle]

QXE = 180o - PXQ = 900 [Linear pair]

In XEQ,

XEQ = 180o - (EXQ + EQX) (Angle sum property)

XEQ = 180o - (900 + 300) = 600

PEQ = 60o

Answered by Expert 4th June 2014, 3:23 PM
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