Question

plz solve ques no.40 41 42 nd 43 do in ur writing on ppr

Answer 1

Heya User,

=_=  I hate the use of paper.. Kindly consider this an answer ...

40 --> 
--> Let CE be external angle bisector ... and BE internal ..

Proof :->
---> ∠ABC + ∠BAC = ∠ACD 
==> ( ∠ABC / 2 ) + ( ∠BAC / 2 ) = ( ∠ACD / 2 )
==> ∠EBC + ( ∠BAC / 2 ) = ∠ECD  --> ( i ) 

However, from exterior angle sum property 0_0 { I'm not sure if that's the ri8 name } 

--> ∠EBC + ∠BEC = ∠ECD --> where ECD is exterior angle --> ( ii )

Hence, from ( i ) and ( ii ) --> BEC = ( ∠BAC / 2 )   ^_^

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41 -> Clearly, OA and OC are angle bisectors 

Now, ∠A + ∠B + ∠C = 180°
===> ( ∠A / 2 ) + ( ∠B / 2 ) + ( ∠C / 2 ) = 90°
===> ( ∠A / 2 ) + [ 180 - ∠AOC ] = 90°
===> ∠AOC = 90² + ( ∠A / 2 ) ..

However, since we have ∠AOC = 135° => ∠A / 2 = 45° => ∠A = 90°
And we are done ^_^

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42 -> By exterior angle sum :->
---> ∠ABC + ∠ACD = ∠ABC + ( ∠ABC + ∠BAC )
                                = ∠ABC + ∠ABC + ∠BAL + ∠CAL
                                = 2 [ ∠ABC + ∠BAL ]
                                = 2 ∠ALC   --->  0_0 Done..

0_0 --> ∠BAL = ∠CAL due to angle bisector ..

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43-> For Qn. 43, I am sorry to say bt I can;t really get what the Qn. is actually asking of meh 

^_^ Sorry for the Delay..