Question

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Answer 1

Question:

The circumference of a circle, with center 0, is divided into three arcs APB, BQC, and CRA such that:

Arc APB/2= Arc BQC/3= Arc CRA/4

Find Angle BOC .

Solution:

From the given conditions given in the question,

We can draw the circle with arc APB, arc BQC, and arc CRA,

The given equation is,

$\frac{arc \: abc}{2} = \frac{arc \: bqc}{3} = \frac{arc \: cra}{4}$

$let \\ \frac{arc \: abc}{2} = \frac{arc \: bqc}{3} = \frac{arc \: cra}{4} = k$

then

Arc APB = 2k, Arc BQC = 3k, Arc CRA = 4k

or

Arc APB : Arc BQC : Arc CRA = 2:3:4

Therefore,

ZAOB : ZBOC: ZAOC = 2:3:4

and also,

and Z AOB = 2k°, ZBOC = 3k°, and ZCRA = 4k°

Now,

Angle in a circle is 360°

So, 2k + 3k + 4k = 360°

9k=360°

k= 40°

Hence, Angle BOC = 3 × 40°= 120°