Question

In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠CBE = 65°, calculate ∠DEC.

Answer 1

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°