Q. 65.0( 1 Vote )

# Two circle drawn

Answer :

∠ COD = 56^{0}

∠ COE = 40^{0}

∠ DAC and ∠ DOC are supplementary to each other since AD and AC are perpendicular to OD and OC respectively

⇒ ∠ DAC = (180^{0}-56^{0}) = 124^{0}

Since DA = AC so Δ DAC is isosceles

∠ ACD = 28^{0}

∠ COE and ∠ CBE are supplementary to each other since BC and BE are perpendicular to OC and OE respectively

⇒ ∠ CBE = (180^{0}-40^{0}) = 140^{0}

Since BC = BE so Δ CBE is isosceles

∠ BCE = 20^{0}

y-x = ∠ ACD-∠ BCE = 8^{0}

∠ CDO = ∠ OCD = 62^{0} (∠ ADO and ∠ ACO is equal to 90^{0})

Hence OD = OC

∠ OEC = ∠ OCE = 70^{0}(∠ OCB and ∠ OEB is equal to 90^{0})

Hence OE = OC

So combining the above two we can say

OD = OC = OE

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