Question

A) The diagram below, not drawn to scale, shows a circle, centre O. The line BC is a tangent to the circle at B. Angle CBD = 42° and angle OBE = 20°.

(i) <BOE
(ii)<OED
(iii)<BFE​

Answer 1

The values of ∠BOE, ∠OED and ∠BFE are 140⁰, 28⁰ and 110⁰ respectively.

How to solve angles in a cyclic quadrilateral.

i) From the figure

OB = OE(radii)

∆OBE is issoceles triangle

∠OBE = ∠BEO(base ∠s of isso. ∆)

∠OBE + ∠BEO + ∠BOE = 180 (sum of int.∠s of ∆)

20 + 20 + ∠BOE = 180

40 + ∠BOE = 180

∠BOE = 180 - 40

∠BOE = 140⁰

ii) OB is a radius and BC is a tangent line with tangency at B.

∠OBC = 90(radius - tangent postulate)

∠OBD = 90 - 42 = 48⁰

∠BOE = 2∠BDE(∠ at center of a circle is twice ∠ subtended at the circumference of the circle)

∠BDE = 1/2 ∠BOE

∠BDE = 1/2*140 = 70⁰

In ∆BDE

∠EBD + ∠BDE + ∠BED = 180( sum of ∠s in a ∆)

(20 + 42) + 70 + ∠BED = 180

132 + ∠BED = 180

∠BED = 180 - 132

∠BED = 48

But

∠BED = ∠BEO + ∠OED

∠OED = 48 - 20

∠OED = 28⁰

iii) ∠BFE + ∠BFE = 180(opp. ∠s of cyclic quadrilateral are supplementary)

∠BFE + 70 = 180

∠BFE = 180 - 70

∠BFE = 110⁰

Therefore, the values of ∠BOE, ∠OED and ∠BFE are 140⁰, 28⁰ and 110⁰ respectively.