Question

math.. circle question.....

Answer 1

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$<b>Answer is proved. Full Explanation: </b>$
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

your answer is these :-

Hope it will help you: -

BC = OB GIVEN
BOC = BCO = Y ( ANGLE OPP. TO EQUAL SIDES ARE EQUAL )

ABO IS AN EXTERIOR ANGLE OF TRIANGLE BOC .

ABO = BOC + BCO .( EXTERIOR ANGLE IS EQUAL TO SUM OF INTERIOR OPP. ANGLE )

= Y + Y = 2Y

OB = OA ( RADII OF SAME CIRCLE )
ABO = BAO
= 2Y
NOW, AOD IS AN EXTERIOR ANGLE OF TRIANGLE AOC..

AOD = BAO + BCO ( EXTERIOR ANGLES IS EQUAL TO SUM OF INTERIOR OPP. ANGLE..

= 2Y + Y = 3Y
X = 3Y..

HOPE YOU UNDERSTAND THIS...
HOPE IT WILL HELP YOU..

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