Question

1. In the given figure (not drawn to scale), BD is
a diameter of the circle with centre 0. C and
A are two points on the circle. BA and CD,
when produced, meet at E. If ZDOC = 60° and
ZABD = 23°, then find ZOBC.
A
E
B
2309
(a) 60°
(b) 30°
(c) 45°
V60°
Find
D
(d) 67°
С
(a) 9​

Answer 1

Answer:

.........

c 45 ............

Step-by-step explanation:

.......

Answer 2

Lets join OC & OD

now in Δ AOC

OA = OC ( Radius)

=> ∠CAO = ∠ACO = x

Simialrly

∠DBO = ∠BDO = y ( as OD = OB = Radius)

ABCD is a cylic Quadrilateral

=> ∠BAC + ∠BDC = 180° ( oposite angles)

∠BAC = ∠OAC as O lies on AB ( as AB is diameter)

∠BDC = ∠BDO + ∠COD

=> ∠OAC + ∠BDO + ∠COD = 180°

=> x + y + ∠COD = 180°

=> ∠COD = 180° - x - y

∠COD = ∠CDO as ( OC = OD = Radius)

=> ∠COD = 180° - x - y

in ΔOCD

∠DCO + ∠COD + ∠DOC= 180°

=> 2 ( 180° - x - y) + 40° = 180°

=> 180° - x - y = 70°

in ΔAEB

∠EAB + ∠EBA + ∠AEB = 180°

∠AEB = ∠CED

=> x + y + ∠CED = 180°

=>∠CED = 180° - x - y

=> ∠CED = 70°