Question

. In the given figure, O is the centre of the circle and
AB is a straight line, then triangle CDO is always alan
(1) Equilateral triangle
(2) Isosceles triangle
(3) Right triangle
​

Answer 1

Given :   O is the centre of the circle and AB is a straight line

To Find : triangle CDO is always a ______

1) isosceles triangle

2) isosceles right triangle

3) equilateral triangle

Solution:

Join AO

=> ∠OAC = ∠OCA = 20°   as OA = OC = Radius

=> ∠AOC = 140°

=> ∠ABC = (1/2)∠AOC = (1/2)140° = 70°

in ΔABC

∠ABC  = 70°

∠BAC  = 30°

=> ∠ACB = 80°

=> ∠BCD = 80° - 20° - 20°  = 40°

∠ABC  = 70° =>  ∠DBC  = 70°

∠BCD =    40°

=> ∠BDC=   70°

Hence ∠DBC  = ∠BDC

=> BC = CD

Join BO  

∠BOC = 2(∠BAC)

=> ∠BOC = 2(30°)

=> ∠BOC = 60°

∠BCO = ∠BCD + ∠DCO = 40° + 20° = 60°

Hence ΔBOC is an equilateral Triangle

=> BC = OC

BC = CD

BC = OC

=> OC = CD

OC = CD hence Δ CDO is an isosceles triangle

∠DCO = 20°  ( hence it can not be equilateral or right angle triangle )  

Δ CDO is always a  isosceles triangle

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