Question

plz solve this 16 th question

Answer 1

BC = OB [GIVEN ]

 BOC = . BCO  = y .     [ ANGLES OPPOSITE TO EQUAL SIDES ARE EQUAL ]

   ABO  IS AN EXTERIOR ANGLE OF TRIANGLE  BOC

. ABO = BOC  + BCO  [EXT. ANGLE  IS EQUAL TO SUM OF  INTERIOR  OPPOSITE ANGLES ]

                    = y + y    =   2y   ...........(1)

OB  =  OA [ RADII OF SAME CIRCLE ]

ABO  = BAO = 2y     [from (1]

now, AOD is an exterior angle of  triangle AOC

   AOD = BAO + BCO  [EXT. ANGLE IS EQUAL TO SUM OF INTERIOR OPPOSITE ANGLES]

           =2y + y  = 3y

  x  =  3y

hence,, proved

Answer 2

given that

IN∆ BOC

BC=BO

BOC= Y°

Exterior angle OBA= y° +y°= 2y°

now in ∆ ABO

AO=OB =2y°

Now In ∆AOC

AOD= OAC+ OCA

X° = y° +2y°

x°= 3y

proved......