Question

In the figure, AB and CD are two straight lines intersecting at 0.
m|AOD=​

Answer 1

$\angle {COD} = 180\degree \: ( Straight \:angle )$

$\implies \angle {COP} + \angle {POB} + \angle {BOD} = 180\degree$

$\implies (3x+7)\degree + (x+5)\degree + 40\degree = 180\degree$

$\implies 3x + 7 + x + 5 + 40 = 180$

$\implies 4x + 52 = 180$

$\implies 4x = 180 - 52$

$\implies 4x = 128$

$\implies x = \frac{128}{4}$

$\implies x = 32\: ---(1)$

$Now , \angle {AOD} = \angle {COB} \\ ( Vertically \: Opposite \: angles )$

$\implies \angle {AOD} = \angle {COP} +\angle {POB}$

$= 3x + 7 + x + 5 \\= 4x + 12 \\= 4 \times 32 + 12 \: [ From \: (1) ]$

$= 128 + 12 \\= 140\degree$

Therefore.,

$\red { Value \: of \:\angle {AOD}} \green {= 140\degree }$

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