Question

(i) In the given figure, AOB is a diameter of the
circle with centre O and ZAOC = 100°, find ZBDC.
A А
100°
B​

Answer 1

Answer:

Angle subtended by chord BC at the centre will be double than that of angle

subtended by it at circumference.

therefore ∠BOC=2∠CAB=2(60)

∘

=120

∘

and ∠CAB=∠CDB=60

∘

{chord substend equal angle on circumference.}

∠COD=180

∘

.

Hence, ∠COB+∠BOD=180

∘

Hence, ∠BOD=180

∘

−120

∘

=60

∘

In triangle BOD, ∠BOD+∠ODB+∠DBO=180

∘

(Angle sum property of a triangle)

Thus, 60

∘

+60

∘

+∠ODB=180

∘

Thus, ∠ODB=60

∘

. Thus, ∠ADB=60

∘

So, option D is the answer