Question

In Fig. 20, OE is the bisector of angle BOD. If angle 1 = 70°, find the magnitudes of angles 2,3,4.

Please answer fast.... ​

Answer 1

$\bold{Given :} \: OE \: is \: the \: bisector \: of \angle \: BOD \: and \: \angle1 = 70 \degree$

$\bold{To \: find} : the \: magnitudes \: of \angle2 \: \angle3 \: \angle4$

$\underline{ \underline{ \bold{Solution}}}$

OE is the bisector of angle DOB

$\implies \: \angle \: doe \: = \angle \: boe \\ \\ \implies \: 70 \degree \: = \angle \: boe \\ \\ \implies \: \angle \: boe \: = \bold{70 \degree} \\ \\ \therefore \: \angle \: dob \: = \angle \: doe \: + \angle \: boe$

$\implies \: \angle \: dob \: = 70 \degree + 70 \degree \\ \\ \implies \: \angle \: dob \: = \underline{\bold{ 140 \degree}}$

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Now

$\angle \: dob = \angle \: aoc \\ (vertically \: opposite \: angle)$

$\implies \: \angle \: aoc \: = \underline{ \bold{140 \degree}}$

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$\angle \: aod + \angle \: dob= 180 \degree \: ( \: linear \: pair)$

$\implies \: \angle \: aod \: = 180 \degree - 140 \degree \\ \\ \implies \: \angle \: aod \: = \underline{ \bold{40 \degree}}$

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And

$\angle \: cob = \angle \: aod \\ ( \: vertically \: opposite \: angles) \\ \\ \implies \: \angle \: cob \: = \underline{ \bold{ \: 40 \degree}}$

Answer 2

$\underline{\huge{\blue{\frak{QuEsTion}}}}$

In the figure , OE is the bisector of angle BOD. If ∠1 = 70°, find the magnitudes of angles 2 , 3 , 4 .

$\underline{\huge{\green{\frak{AnswEr}}}}$

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∵ OE is the bisector of ∠BOD

∴ ∠ BOD = 2 ∠1 = 2(70°) = 140°

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∵ COD is line

∴ ∠BOD + ∠4 = 180° (linear pair)

140° + ∠ 4 = 180°

∠ 4 = 180° - 140°

∠ 4 = 40°

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∵ ∠ 3 and ∠ BOD are vertically opposite angles

∴ ∠ 3 = ∠ BOD = 140°

∠ 3 = 140°

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∵ ∠2 and ∠4 are vertically opposite angles

∴ ∠ 2 = ∠4 = 40°

∠ 2 = 40°

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