In figure, points G, D, E, F are concyclic points of a circle with centre C.
$∠ E C F = 70^{o}$ , m (arc DGF) = $200^{o}$ find m (arc DE) and m(arc DEF).

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Solution

$∠ E C F = 70^{o} = m (a r c E F) = 70^{o}$
$m (a r c D G F) = 200^{o}$
$m (a r c D G F) + m (a r c E F) + m (a r c D E) = 360^{o}$
$200 + 70 + m (a r c D E) = 360^{o}$
$m (a r c D E) = 90^{o}$
$m (a r c D E F) = m (a r c D E) + m (a r c E F)$
$m (a r c D E F) = 90 + 70 = 160^{o}$

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In figure, points G, D, E, F are concyclic points of a circle with centre C. ∠ECF=70o, m (arc DGF) = 200o find m (arc DE) and m(arc DEF).

∠ECF=70o=m(arcEF)=70om(arcDGF)=200om(arcDGF)+m(arcEF)+m(arcDE)=360o200+70+m(arcDE)=360om(arcDE)=90om(arcDEF)=m(arcDE)+m(arcEF)m(arcDEF)=90+70=160o

In figure, points G, D, E, F are concyclic points of a circle with centre C.
$∠ E C F = 70^{o}$ , m (arc DGF) = $200^{o}$ find m (arc DE) and m(arc DEF).

$∠ E C F = 70^{o} = m (a r c E F) = 70^{o}$
$m (a r c D G F) = 200^{o}$
$m (a r c D G F) + m (a r c E F) + m (a r c D E) = 360^{o}$
$200 + 70 + m (a r c D E) = 360^{o}$
$m (a r c D E) = 90^{o}$
$m (a r c D E F) = m (a r c D E) + m (a r c E F)$
$m (a r c D E F) = 90 + 70 = 160^{o}$