is a diameter of the circle and C, D, E are any three points on the se

1.	is a diameter of the circle and C, D, E are any three points on the semi-circle, then find theIf ABvalue of ZACDLBED.Ahetween their centres is 4
Answer» AOB is a diameter of the circle, C, E, D are three points on the circle To find:∠ACD +∠BED Construction: Join CO, DO and EO Assume AC = CD = AO Assume DE = EB So inΔACO It is an equilateraltriangle∠ACO = 60° InΔCDO It is an equilateral triangle, hence∠DCO = 60° ∠DOE =∠EOB As∠AOB = 180° So 180° =∠AOC +∠COD +∠DOE +∠EOB And∠AOC =∠COD = 60° And∠DOE =∠EOB Hence 2∠DOE = 180° - 60° - 60° ∠DOE= 30° InΔODE ∠ODE =∠OED (isoscelestriangle) And∠ODE +∠OED +∠DOE = 180° So 2∠ODE = 180° - 30° = 150° So∠ODE = 75° InΔOEB ∠OEB =∠OBE (isosceles triangle) And∠OEB +∠OEB +∠BOE = 180° So 2∠OEB = 180° - 30° = 150° So∠OEB = 75° So∠ACD +∠BED =∠ACO +∠DCO +∠DEO +∠BEO = 60° + 60° + 75° + 75° = 270°

is a diameter of the circle and C, D, E are any three points on the semi-circle, then find theIf ABvalue of ZACDLBED.Ahetween their centres is 4

Answer»

AOB is a diameter of the circle, C, E, D are three points on the circle

To find:∠ACD +∠BED

Construction: Join CO, DO and EO

Assume AC = CD = AO

Assume DE = EB

So inΔACO

It is an equilateraltriangle∠ACO = 60°

InΔCDO

It is an equilateral triangle, hence∠DCO = 60°

∠DOE =∠EOB

As∠AOB = 180°

So 180° =∠AOC +∠COD +∠DOE +∠EOB

And∠AOC =∠COD = 60°

And∠DOE =∠EOB

Hence 2∠DOE = 180° - 60° - 60°

∠DOE= 30°

InΔODE

∠ODE =∠OED (isoscelestriangle)

And∠ODE +∠OED +∠DOE = 180°

So 2∠ODE = 180° - 30° = 150°

So∠ODE = 75°

InΔOEB

∠OEB =∠OBE (isosceles triangle)

And∠OEB +∠OEB +∠BOE = 180°

So 2∠OEB = 180° - 30° = 150°

So∠OEB = 75°

So∠ACD +∠BED =∠ACO +∠DCO +∠DEO +∠BEO = 60° + 60° + 75° + 75° = 270°