Question

In the given figure, O is the centre of the circle and AB is the diameter of the circle. If difference of ∠COD and ∠DBC is 25°, then find the measure of ∠DEB.

Answer 1

In C(O, r),

• AB = Diameter.
• ∠COD - ∠DBC = 25°.

Since AB is the diameter of C(O, r), we can say that arc ADCB is a semi-circle.

⟹ ∠ADB = 90° [Angle subtended by a semi-circle]

or, ∠EDB = 90° [since BD becomes perpendicular to AE]

Let us join CD such that it becomes a chord.

It is given that,

∠COD - ∠DBC = 25°

⟹ 2∠DBC - ∠DBC = 25°

[since angle subtended by a chord at centre is double the angle subtended by any other point of the circle.]

⟹ ∠DBC = 25°

We know,

Sum of interior angles of a △ = 180°

In △BDE,

∴ ∠DBC + ∠EDB + ∠DEB = 180°

⟹ 25° + 90° + ∠DEB = 180°

⟹ 115° + ∠DEB = 180°

⟹ ∠DEB = 180° - 115°

⟹ ∠DEB = 65°.