The circumference of a circle with centre O is divided into three arcs APB, BQC and CRA such that

$\frac{a r c A P B}{2} = \frac{a r c B Q C}{3} = \frac{a r c C R A}{4}$ . $∠$ BOC = ___ degrees

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Solution

Let $\frac{a r c A P B}{2} = \frac{a r c B Q C}{3} = \frac{a r c C R A}{4} = k$ .

$\Rightarrow a r c A P B = 2 k, a r c B Q C = 3 k, C R A = 4 k$ .

$\Rightarrow$ arc APB : arc BQC : arc CRA = 2 : 3 : 4

$\Rightarrow ∠ A O B : ∠ B O C : ∠ C O A$ = 2 : 3 : 4

$∴ ∠ B O C = \frac{3}{9} \times 360^{o}$

$\Rightarrow ∠ B O C = 120^{o}$

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The circumference of a circle with centre O is divided into three arcs APB, BQC and CRA such that arc APB2 = arc BQC3 = arc CRA4. ∠ BOC = ___ degrees

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The circumference of a circle with centre O is divided into three arcs APB, BQC and CRA such that

$\frac{a r c A P B}{2} = \frac{a r c B Q C}{3} = \frac{a r c C R A}{4}$ . $∠$ BOC = ___ degrees