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The base B C of triangle A B C is produc...

The base `B C` of triangle `A B C` is produced both ways and the measure of exterior angles formed are `94^0a n d\ 126^0dot` Then `/_B A C=` `94^0` (b) `54^0` (c) `40^0` (d) `44^0`

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Given: `/_ABD=94°` and `/_ACE=126°`
Solution: `/_ABD+/_ABC=180°`(Linear pair)
`94°+/_ABC=180°`
`/_ABC=180°-94°=86°`
`/_ACE+/_ACB=180°`(Linear pair)
`=>126°+/_ACB=180°`
`=>/_ACB=180°-126°=54°`
In `/_\ABC`,
`/_ABC+/_ACB+/_BAC=180°`
`=>86°+54°+/_BAC=180°`
`=>/_BAC=180°-140°`
`=>/_BAC=40°`
`therefore /_BAC=40°`
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