emzogwayo
Answered

Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Our platform connects you with professionals ready to provide precise answers to all your questions in various areas of expertise. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

In the figure below AOD is a diameter of the circle cetre O. BC is a chord parallel to AD. FE is a tangent to the circle. OF bisects angle COD. Angle BCE = angle COE = 20° BC cuts OE at X
Calculate;
(a) angle BOE
(b) angle BEC
(c) angle CEF
(d) angle OXC
(e) angle OFE​

In The Figure Below AOD Is A Diameter Of The Circle Cetre O BC Is A Chord Parallel To AD FE Is A Tangent To The Circle OF Bisects Angle COD Angle BCE Angle COE class=

Sagot :

mtngkhoi

Step-by-step explanation:

∠ACB=90∘

[∠ from diameter]

In ΔACB

∠A+∠ACB+∠CBA=180∘

∠CBA=180∘

−(90+30)

∠CBA=60∘ (1)

In △OCB

OC=OB

So, ∠OCB=∠OBC        

[The sides are equal]

∠OCB=60∘

∠OCD=90∘

∠OCB+∠BCD=90°

∠BCD=30∘ (2)

∠CBO=∠BCO+∠CDB    

[external ∠ bisectors]

60=30+∠CDB

∠CDB=30° (3)

from (2) & (3)

BC=BD    

[The ∠.S are equal]

View image mtngkhoi
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.