Question

In the given figure, BC is tangent to the circle at point B of circle centred at O. BD is a chord of the circle so that ∠BAD = 55°. Find m∠DBC.

Answer 1

In ΔADB,

∠ADB = 90°  (Angle in semi-circle) ...(i)

Now, by using angle sum properly in ΔABD, we have

∠ABD + ∠ADB + ∠DAB = 180°

⇒ ∠ABD + 90° + 55° = 180°  ...[From equation (i) and given ∠DAB = 55°]

⇒ ∠ABD = 180° – 145°

⇒ ∠ABD = 35°  ...(ii)

Now, ∠ABD = 90°  ...(Angle between tangent and radius)

or, ∠ABD + ∠DBC = 90°

or, ∠DBC = 90° – ∠ABD

or, ∠DBC = 90° – 35°   ...[From equation (ii)]

or, ∠DBC = 55°

Alternate Method: Since, the angle between chord and tangent is equal to the angle subtended by the same chord in alternate segment of the circle, Hence, ∠BDC = ∠BAD = 55°.