Question

 AB and CD are two straight lines which intersect at point O. ∠AOD and ∠COB are vertically opposite to each other. If ∠AOC is 80°, find the value of ∠AOD and ∠DOB​

Answer 1

Let ∠AOD=4x and ∠DOB=5x

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘ ⇒x=

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘ ⇒x= 9

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘ ⇒x= 9180

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘ ⇒x= 9180 ∘

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘ ⇒x= 9180 ∘

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘ ⇒x= 9180 ∘

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘ ⇒x= 9180 ∘ =20

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘ ⇒x= 9180 ∘ =20 ∘

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘ ⇒x= 9180 ∘ =20 ∘

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘ ⇒x= 9180 ∘ =20 ∘ ∠AOD=4x

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘ ⇒x= 9180 ∘ =20 ∘ ∠AOD=4x⇒∠AOD=4×20

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘ ⇒x= 9180 ∘ =20 ∘ ∠AOD=4x⇒∠AOD=4×20 ∘

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘ ⇒x= 9180 ∘ =20 ∘ ∠AOD=4x⇒∠AOD=4×20 ∘ =80

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘ ⇒x= 9180 ∘ =20 ∘ ∠AOD=4x⇒∠AOD=4×20 ∘ =80 ∘

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘ ⇒x= 9180 ∘ =20 ∘ ∠AOD=4x⇒∠AOD=4×20 ∘ =80 ∘

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘ ⇒x= 9180 ∘ =20 ∘ ∠AOD=4x⇒∠AOD=4×20 ∘ =80 ∘ Now ∠COB=∠AOD (vertically opposite angles)

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘ ⇒x= 9180 ∘ =20 ∘ ∠AOD=4x⇒∠AOD=4×20 ∘ =80 ∘ Now ∠COB=∠AOD (vertically opposite angles)⇒∠COB=80

Let ∠AOD=4x and ∠DOB=5x∠AOD+∠DOB=180 ∘ (Adjacent angle on straight line are supplementary)⇒4x+5x=180 ∘ ⇒9x=180 ∘ ⇒x= 9180 ∘ =20 ∘ ∠AOD=4x⇒∠AOD=4×20 ∘ =80 ∘ Now ∠COB=∠AOD (vertically opposite angles)⇒∠COB=80 ∘

hope it helps you ✌️✅✅✅

Answer 2

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