Câu 21 trang 102 Sách Bài Tập (SBT) Toán 9 Tập 2

Hãy tính các góc của tam giác DEF.

Cho tam giác ABC nội tiếp trong đường tròn tâm O, biết (widehat A = {32^0}), (widehat B = {84^0}). Lấy các điểm D, E, F thuộc đường tròn tâm O sao cho AD = AB, BE = BC, CF = CA. Hãy tính các góc của tam giác DEF.

Giải

 

(widehat A = {1 over 2})  sđ (overparen{BC}) (tính chất góc nội tiếp)

( Rightarrow ) sđ (overparen{BC}) ( = 2widehat A = {2.32^0} = {64^0})

BC = BE (gt)

( Rightarrow ) sđ (overparen{BC}) = sđ (overparen{BE}) = 64⁰

(widehat B = {1 over 2}) sđ (overparen{AC}) (tính chất góc nội tiếp)

( Rightarrow ) sđ (overparen{AC}) ( = 2widehat B = {2.84^0} = {168^0})

AC = CF (gt)

( Rightarrow ) sđ (overparen{CF}) =  sđ (overparen{AC}) = 168⁰

 sđ (overparen{AC}) +  sđ (overparen{AF}) +  sđ (overparen{CF}) = 360⁰

( Rightarrow ) sđ (overparen{AF}) ( = {360^0} - )  sđ (overparen{AC}) -  sđ (overparen{CF}) = 360⁰ – 168⁰. 2 = 24⁰

Trong ∆ABC ta có: (widehat A + widehat B + widehat C = {180^0})

( Rightarrow widehat {ACB} = {180^0} - left( {widehat A + widehat B} ight))

              = ({180^0} - left( {{{32}^0} + {{84}^0}} ight) = {64^0})

 sđ (widehat {ACB} = {1 over 2}) sđ (overparen{AB})

( Rightarrow ) sđ (overparen{AB}) ( = 2widehat {ACB} = {2.64^0} = {128^0})

AD = AB (gt)

( Rightarrow ) sđ (overparen{AD}) =  sđ (overparen{AB}) = 128⁰

(widehat {FED} = {1 over 2}) sđ (overparen{DF}) ( = {1 over 2}) ( sđ (overparen{AD}) +  sđ (overparen{AF}))

            = ({1 over 2}.left( {{{128}^0} + {{24}^0}} ight) = {76^0})

(widehat {EDF} = {1 over 2}) sđ (overparen{EF}) = ({1 over 2}) ( sđ (overparen{AB}) -  sđ (overparen{AF}) -  sđ (overparen{BE})

            = ({1 over 2}.left( {{{128}^0} - {{24}^0} - {{64}^0}} ight) = {20^0})

(widehat {DFE} = {180^0} - left( {widehat {FED} + widehat {EDF}} ight))

            = ({180^0} - left( {{{76}^0} + {{20}^0}} ight) = {84^0}).

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