Ta có :
`hat( B) + hat( C) =` $180^o$
`hat( 3C) =` $180^o$
`⇒ hat( C) = 180 :3 =` $60^o$
`⇒ hat( B) = hat( 2C) = 2 xx 60 =` $120^o$
`=> hat( A) = hat( D) =` $40^o$
Vì `hat( A) + hat( D) =` $180^o$
`⇒ hat( D) +` $40^o$ `+ hat( D) =` $180^o$
`⇒ hat( 2D) +` $40^o$ `=` $180^o$
`⇒ hat( 2D) =` $140^o$
`⇒ hat( D) =` ${ 140 }/{ 2 }$ `=` $70 ^o$
`=>` `hat( A) = hat( D) +` $40^o$
`⇒ hat( A) =` $40^o$ + $70^o$
`⇒ hat( A) =` $110 ^o$
`=>hat( B) =` $120^o$
`=> hat( C) =` $60^o$
`=>hat( D) =` $70 ^o$
