Math, asked by somanshu2basa, 19 days ago

Find the value of ‘p and also find the measure of  BOC and  AOD.

Answers

Answered by Kaushalsingh74883508
1

Step-by-step explanation:

We know that the sum of all the angles in a straight line is 180

We know that the sum of all the angles in a straight line is 180 o

We know that the sum of all the angles in a straight line is 180 o

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50 o

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50 o

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50 o ∠COD=50

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50 o ∠COD=50 o

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50 o ∠COD=50 o

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50 o ∠COD=50 o ∠AOD=x+10

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50 o ∠COD=50 o ∠AOD=x+10 o

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50 o ∠COD=50 o ∠AOD=x+10 o

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50 o ∠COD=50 o ∠AOD=x+10 o =60

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50 o ∠COD=50 o ∠AOD=x+10 o =60 o

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50 o ∠COD=50 o ∠AOD=x+10 o =60 o

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50 o ∠COD=50 o ∠AOD=x+10 o =60 o ∠BOC=x+20=70

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50 o ∠COD=50 o ∠AOD=x+10 o =60 o ∠BOC=x+20=70 o

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50 o ∠COD=50 o ∠AOD=x+10 o =60 o ∠BOC=x+20=70 o

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50 o ∠COD=50 o ∠AOD=x+10 o =60 o ∠BOC=x+20=70 o ∴ ∠COD=50

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50 o ∠COD=50 o ∠AOD=x+10 o =60 o ∠BOC=x+20=70 o ∴ ∠COD=50 o

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50 o ∠COD=50 o ∠AOD=x+10 o =60 o ∠BOC=x+20=70 o ∴ ∠COD=50 o , ∠AOD=60

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50 o ∠COD=50 o ∠AOD=x+10 o =60 o ∠BOC=x+20=70 o ∴ ∠COD=50 o , ∠AOD=60 oand ∠BOC=70

We know that the sum of all the angles in a straight line is 180 o ∠BOC+∠COD+∠AOD=180 o .x+20+x+x+10=180 o 3x+30=180 o 3x=150 o x=50 o ∠COD=50 o ∠AOD=x+10 o =60 o ∠BOC=x+20=70 o ∴ ∠COD=50 o , ∠AOD=60 oand ∠BOC=70 o.

Similar questions