The circumference of a circle with centre o is divided into three arcs apb, bqc, cra such that: arc apb ÷2 = arc bqc÷3=arc cra ÷4.. Find boc
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Answer:
∠BOC = 120°
Step-by-step explanation:
Here, It is given that arc apb ÷2 = arc bqc÷3=arc cra ÷4
=> (arc apb)/2 = (arc bqc/3) = (arc cra/4)
Let this is equal to k .
=> (arc apb)/2 = (arc bqc/3) = (arc cra/4) = k
=> (arc apb) = 2k
=> (arc bqc) = 3k
=> (arc cra) = 4k
If we add the three arcs the complete circle is formed .
So, (arc apb) + (arc bqc) + (arc cra) = 2k + 3k + 4k
= 9k
This is equal to circumference of circle .
=> 2pi * r = 9k
As, arc of circle = (∅/360) * (2pi * r)
where, ∅ = angle subtended by arc .
Here, ∠BOC is the angle subtended by arc BQC .
=> 3k = (∅/360) * (2pi * r)
=> 3k = (∅/360) * 9k
=> ∅ = 120°
Thus, ∠BOC = 120°
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