plzzzz frds help me... plzzzz
Attachments:
Answers
Answered by
0
ok I help you later sorry I am busy
Answered by
0
<DOA = x
<BCO = y
OD=OA=OB=BC
∆ADO, ∆ABO & ∆BOC are isosceles ∆le
So, <BCO = <BOC = y
<OBC = 180°- 2y
<ABO = <OAB = 180° - (180°- 2y) = 2y
<AOB = 180° - (2y+2y) = 180° - 4y
<DOA + <AOB + <BOC = 180°
x + (180° - 4y) + y = 180°
x + 180° -4y + y = 180°
x = 3y
Answer is 3
<BCO = y
OD=OA=OB=BC
∆ADO, ∆ABO & ∆BOC are isosceles ∆le
So, <BCO = <BOC = y
<OBC = 180°- 2y
<ABO = <OAB = 180° - (180°- 2y) = 2y
<AOB = 180° - (2y+2y) = 180° - 4y
<DOA + <AOB + <BOC = 180°
x + (180° - 4y) + y = 180°
x + 180° -4y + y = 180°
x = 3y
Answer is 3
Similar questions