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Question
In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠CBE = 65°, calculate ∠DEC.
Solution
Join OE,
Arc EC subtends ∠EOC at the centre and ∠EBC at the remaining part of the circle.
∠EOC = 2∠EBC = 2 × 65° = 130°
Now in ΔOEC, OE = OC ...[Radii of the same circle]
∴ ∠OEC = ∠OCE
But, in ΔEOC,
∠OEC + ∠OCE + ∠EOC = 180° ...[Angles of a triangle]
`=>` ∠OCE + ∠OCE + ∠EOC = 180°
`=>` 2∠OCE + 130° = 180°
`=>` 2∠OCE = 180° – 130°
`=>` 2∠OCE = 50°
`=> ∠OCE = 50^circ/2 = 25^circ`
∴ AC || ED ...[Given]
∴ ∠DEC = ∠OCE
`=>` ∠DEC = 25°
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