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 bis
ects 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
1 answer:
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]
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