Cho tam giac ABC, biet B=50do, C=30 do
Tinh cac goc ngoai tai dinh A,B,C; Ve hinh minh hoa
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Cho tam giac ABC .Cac duong thang chua tia phan giac cua cac goc ngoai dinh B va o dinh C cat nhau tai K.Tinh goc a biet goc BCK =50°
1.cho tam giac ABC co goc B-C=alpha phan giac AD
a) tinh gocADC,ADB theo alpha
b)Ve duong cao AH tinh goc HAD
2.chotam giac ABC,gocA =alpha,cac tia ohan giac cua B va C cat nhay tai I.Phan giac cat goc ngoai tai dinh B,C cat nhau tai K.Tia p/g goc B cat goc ngoai o dinh C.tinh cac goc BIC,BKC,BEC theo alpha.
cho tam giac abc can tai a co goc bac =50do tren tia doi cua tia bc lay diem d tren tia doi cua tia cb lay diem e sao cho bd =ba ce=ca tinh goc dae
cho tam giac abc deu ve ben ngoai tam giac cac tam giac abd vuong can tai b tam giac ace vuong can tai c tinh so goc nhon cua ade
XÉT \(\Delta ABC\)CÂN TẠI A
\(\Rightarrow\hept{\begin{cases}AB=AC\\\widehat{B}=\widehat{C}\end{cases}}\)
TA CÓ \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\left(Đ/L\right)\)
THAY\(50^0+\widehat{B}+\widehat{C}=180^o\)
\(\widehat{B}+\widehat{C}=130^o\)
MÀ\(\widehat{B}=\widehat{C}\)
\(\Rightarrow\widehat{B}=\widehat{C}=\frac{130^o}{2}=65^o\)
TA CÓ \(\widehat{DBA}+\widehat{ABC}=180^o\left(KB\right)\)
\(\Rightarrow\widehat{DBA}=180^o-65^o=115^o\)
TA CÓ\(\widehat{ACE}+\widehat{ACB}=180^o\left(KB\right)\)
\(\Rightarrow\widehat{ACE}=180^o-65^0=115^o\)
XÉT \(\Delta ACE\)CÓ AC=CE (GT) =>\(\Delta ACE\)CÂN TẠI C
\(\Rightarrow\widehat{CAE}=\widehat{AEC}=\frac{180^o-115^0}{2}=32,5^0\)
XÉT \(\Delta ABD\)CÓ AB=BD (GT) =>\(\Delta ABD\)CÂN TẠI B
\(\Rightarrow\widehat{DAB}=\widehat{ADB}=\frac{180^o-115^0}{2}=32,5^0\)
TA CÓ\(\widehat{DAB}+\widehat{BAC}+\widehat{EAC}=\widehat{DAE}\)
THAY\(32,5^o+50^0+32,5^0=\widehat{DAE}\)
\(\Rightarrow\widehat{DAE}=115^0\)
cho tam giac ABC co B=110 do C =50 do phan giac goc ngoai tai dinh A cat tia doi cua tia BC tai D tinh cac goc cua tam giac ABD
cho hinh thang ABCD day AB, CD cac duong phan giac cac goc ngoai tai dinh A, D cat nhau o M . Cac duong phan giac goc ngoai tai dinh B, C cat nhau o N
a/ chung minh MN//CD
b/ tinh chu vi hinh thang ABCD biet MN=4cm
cho tu giac abcd biet goc a: goc b: goc c: goc d =4:3:2:1
a) tinh cac goc cua tu giac abcd
b) cac tia phan giac cua goc c va goc d cat nhau tai e. cac duong phan giac cua goc ngoai tai ca dinh c va d cat nhau tai f. tinh goc ced va goc cfd
a) Ta có: \(\frac{\widehat{A}}{4}=\frac{\widehat{B}}{3}=\frac{\widehat{C}}{2}=\frac{\widehat{D}}{1}\)
Áp dụng t/c dãy tỉ số bằng nhau ta được:
\(\frac{\widehat{A}}{4}=\frac{\widehat{B}}{3}=\frac{\widehat{C}}{2}=\frac{\widehat{D}}{1}=\frac{\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}}{4+3+2+1}=\frac{360}{10}=36\)
\(\Rightarrow\widehat{A}=144^0;\widehat{B}=108^0;\widehat{C}=72^0;\widehat{D}=36^0\)
hai tia phan giac trong tai dinh b va c cua tam giac abc cat nhau tai o, biet goc boc bang 130 .
a,tinh so do goc a
b,hai tia phan giac ngoai tai dinh b va c cua tam giac abc cast nhau tai p.chung minh a,o,p thang hang
c
tam giac abc la tam gia i
cho tam giac ABC co goc A=a . Cac tia phan giac cua goc B va C cat nhau tai I . Cac tia phan giac cua cac goc ngoai ding B va C cat nhau tai K. Tia phan giac cua g oc B cat tia phan giac ngoai ding C tai E. Tinh so do cua BKC, B . C theo a
cho tam giac ABC.cac duong thang chua tia phan giac cua goc ngoai o dinh B va o dinh C cat nhau tai K.Tinh goc A biet goc BKC = 500