cho tam giac abc vuong oa co canh BC=2AB. tinh so do cac goc cua tam giac
mau len mk dg can gap
cho tam giac abc vuong oa co canh BC=2AB. tinh so do cac goc cua tam giac
cho tam giac abc vuong oa co canh BC=2AB. tinh so do cac goc cua tam giac
Ai lm dc cho 2 lik lun
|x-1,25|=2,75
Cho tam giac ABC co B=45°,C=75°. Phan giac BAC cat canh BC tai M. Qua M ke dg thang song song voi dg thang AB cat canh AC tai N.
1, tinh so do cac goc BAC va goc AMB
2,tinh so do cac goc cua tam giac AMN
Bài làm
\(\text{| x - 1,25 | = 2,75}\)
\(\Rightarrow\hept{\begin{cases}x-1,25=2,75\Rightarrow x=4\\x-1,25=-2,75\Rightarrow x=-1,5\end{cases}}\)
Vậy x = 4 hoặc x = -1,5
Cho tam giac ABC can tai A co AD la duong trung tuyen
a)Chung minh tam giac ABD= tam gaic ACD va AD vuong goc voi BC
b)Cho AB=10cm,BC=16cm. Tinh do dai AD va so sanh cac goc cua tam giac ABC.
c) Ve duong trung tuyen CF cua tam giac ABC cat AD tai M. Tinh do dai AM.
d) Ve DH vuong goc AC tai H, tren canh AC va canh DC lan luot lay hai diem E,K sao cho AE=AD va DK=DH. Chung minh: EK vuong goc voi BC
A,
xét \(\Delta ABD\)và \(\Delta ACD\)
CÓ \(\hept{\begin{cases}AB=AC\\chungAD\\BD=DC\end{cases}}\)
SUY RA \(\Delta ABD\)=\(\Delta ACD\) (C.C.C) (1)
=> \(\widehat{BDA}\)=\(\widehat{CDA}\)
MÀ \(\widehat{BDA}\)+\(\widehat{CDA}\)=180
=> \(\widehat{BDA}\)=\(\widehat{CDA}\)=90
B, (1) => BC=DC=1/2 BC=8
ÁP DỤNG ĐỊNH LÍ PITAGO TA CÓ
\(AB^2=AD^2+BD^2\)
=> AD^2=36
=>AD=6
cho tam giac can ABC (AB=AC) co goc A khong phai goc vuong . Tu A ke mot tia vuong goc voi AB cat duong thang BC o D. TInh hieu giua cac so do cua hai goc C va D cua tam giac ACD
a)C/M neu tam giac vuong co mot canh goc vuong bang nua canh huyen thi goc doi dien voi canh ay bang 30'
b)Cho tam giac ABC, goi M la trung diem cua BC, ve AH vuong goc voi BC (H thuoc canh BC). Biet rang BAH=HAM=MAC. Tinh cac goc cua tam giac ABC.
cho tam giac ABC co goc A<90 do ve ra ngoaitam giac do hai doan thang AD vuong goc va bang AB, AE vuong goc va bang AC
a, chung minh DC vuong goc va bang BE
b, goi N la trung diem cua DE.Tren tia doi cua NA lay M sao cho NA=NM
chung minh AB=ME va tam giac ABC=tam giac EMA
c, chung minh MA vuong goc voi BC
giup mk cau c thoi cung dc mk dg can gap nha
A=1/4/23-5/21-4/23+1,5+1/16/12
Cho tam giac ABC co B=45°,C=75°.phan giac BAC cat canh BC tai M. Qua M ke dg thang song song voi dg thang AB cat canh AC tai N
1, tinh so do goc BAC va goc AMB
2, tinh so do cac goc cua tam giac AMN
Ve hinh dum nha
giai trước nè
a)Ta có: BAC + ABC + ACB = 180o
BAC = 1800 - ABC - ACB
BAC = 180o -45o- 75o
BAC = 60o
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\(32,5^o+50^0+32,5^0=\widehat{DAE}\)
\(\Rightarrow\widehat{DAE}=115^0\)