Question

In the given figure, AB is a diameter of a circle with centre O. If ADE and CBE are straight lines, meeting at E such that ∠BAD = 35° and ∠BED = 25°, find
(i) ∠DBC
(ii) ∠DCB
(iii) ∠BDC.

Answer 1

AB is a diameter of the circle with centre O.
ADE and CBE are straight lines that meet at E such that ∠BAD = 35° and ∠BED = 25°.
Join BD and AC.

(i)
Now, ∠BDA = 90° (Angle in the semicircle)
Also, ∠EDB + ∠BDA = 180° (Linear pair)
Or ∠EDB = 90°

Now, ∠EBD = {180° – (∠EDB + ∠BED} (Angle sum property)
= (180° – (90° + 25°)
= (180° – 115°) = 65°
∠DBC = (180° – ∠EBD) = (180° - 65°) = 115° (Linear pair)
∴ ∠DBC = 115°

(ii)
Here, ∠DCB = ∠BAD (Angles in the same segment)
∠BAD = 35°
∴ ∠DCB = 35°
(iii)
∠BDC = 180° – (∠DBC + ∠DCB) (Angle sum property)
= {180° – (∠DBC + ∠BAD)}
= 180° – (115° + 35°)
=180° – 150°
= 30°
∴ ∠BDC = 30°