Bài 3: Đặt \(\hat{A}=a;\hat{B}=b;\hat{C}=c\)

Xét ΔABC có \(\hat{A}+\hat{B}+\hat{C}=180^0\)

=>a+b+c=180

Ta có: \(\hat{C}-3\cdot\hat{B}-2\cdot\hat{A}=-3^0\)

=>c-3b-2a=-3

=>2a+3b-c=3

mà a+b+c=180

nên 2a+3b-c+a+b+c=3+180

=>3a+4b=183

=>6a+8b=366

\(5\cdot\hat{B}-2\cdot\hat{A}=16^0\)

=>5b-2a=16

=>15b-6a=48

=>15b-6a+6a+8b=366+48

=>23b=414

=>\(b=\frac{414}{23}=18^0\)

=>\(\hat{B}=18^0\)

3a+4b=183

=>3a=183-4b=183-72=111

=>\(a=\frac{111}{3}=37^0\)

=>\(\hat{A}=37^0\)

\(\hat{C}=180^0-18^0-37^0=180^0-55^0=125^0\)

Bài 2:

Đặt \(\hat{A}=a;\hat{B}=b;\hat{C}=c\)

Xét ΔABC có \(\hat{A}+\hat{B}+\hat{C}=180^0\)

=>a+b+c=180

\(\hat{A}+\hat{B}-2\cdot\hat{C}=27^0\)

=>a+b-2c=27

=>(a+b+c)-(a+b-2c)=180-27

=>3c=153

=>\(c=\frac{153}{3}=51\)

=>\(\hat{C}=51^0\)

\(\hat{A}+3\cdot\hat{C}=273^0\)

=>\(\hat{A}=273^0-3\cdot51^0=273^0-153^0=120^0\)

\(\hat{B}=180^0-51^0-120^0=60^0-51^0=9^0\)

bài 1:

Đặt \(\hat{A}=a;\hat{B}=b;\hat{C}=c\)

Xét ΔABC có \(\hat{A}+\hat{B}+\hat{C}=180^0\)

=>a+b+c=180

\(\hat{A}-\hat{B}+\hat{C}=90^0\)

=>a-b+c=90

=>a+b+c-(a-b+c)=180-90

=>2b=90

=>b=45

=>\(\hat{B}=45^0\)

=>\(\hat{A}+\hat{C}=180^0-45^0=135^0\)

mà \(\hat{A}-\hat{C}=-5^0\)

nên \(\hat{A}=\frac{135^0-5^0}{2}=\frac{130^0}{2}=65^0\)

=>\(\hat{C}=135^0-65^0=70^0\)

a: \(\hat{HAB}+\hat{HAC}\)

\(=\hat{HAB}+\hat{HAB}+\hat{BAC}=2\cdot\hat{HAB}+2\cdot\hat{BAD}\)

\(=2\left(\hat{HAB}+\hat{ABD}\right)=2\cdot\hat{HAD}\)

b: Xét ΔBHA có \(\hat{ABC}\) là góc ngoài tại đỉnh B

nên \(\hat{ABC}=\hat{BHA}+\hat{BAH}=90^0+\hat{BAH}\)

ΔHAC vuông tại H

=>\(\hat{HAC}+\hat{HCA}=90^0\)

=>\(\hat{HCA}=90^0-\hat{HAC}\overline{}\)

c: \(\frac12\cdot\left(\hat{ABC}-\hat{ACB}\right)=\frac12\left(90^0+\hat{HAB}-90^0+\hat{HAC}\right)\)

\(=\frac12\cdot\left(\hat{HAB}+\hat{HAC}\right)=\frac12\cdot2\cdot\hat{HAD}=\hat{HAD}\)

a: \(\hat{HAB}+\hat{HAC}\)

\(=\hat{HAB}+\hat{HAB}+\hat{BAC}=2\cdot\hat{HAB}+2\cdot\hat{BAD}\)

\(=2\left(\hat{HAB}+\hat{ABD}\right)=2\cdot\hat{HAD}\)

b: Xét ΔBHA có \(\hat{ABC}\) là góc ngoài tại đỉnh B

nên \(\hat{ABC}=\hat{BHA}+\hat{BAH}=90^0+\hat{BAH}\)

ΔHAC vuông tại H

=>\(\hat{HAC}+\hat{HCA}=90^0\)

=>\(\hat{HCA}=90^0-\hat{HAC}\overline{}\)

c: \(\frac12\cdot\left(\hat{ABC}-\hat{ACB}\right)=\frac12\left(90^0+\hat{HAB}-90^0+\hat{HAC}\right)\)

\(=\frac12\cdot\left(\hat{HAB}+\hat{HAC}\right)=\frac12\cdot2\cdot\hat{HAD}=\hat{HAD}\)

a, Theo định lí Pytago tam giác ABC vuông tại A

\(BC=\sqrt{AB^2+AC^2}=5cm\)

Theo định lí Pytago tam giác MNP vuông tại N

\(NP=\sqrt{MP^2-MN^2}=6cm\)

b, Xét tam giác ABC và tam giác NPM có 

^BAC = ^PNM = 90⁰

\(\dfrac{AB}{NP}=\dfrac{AC}{NM}=\dfrac{3}{6}=\dfrac{4}{8}=\dfrac{1}{2}\)

Vậy tam giác ABC ~ tam giác NPM ( c.g.c ) 

a: \(BC=\sqrt{AB^2+AC^2}=5\left(cm\right)\)

\(NP=\sqrt{10^2-8^2}=6\left(cm\right)\)

b: Xét ΔABC vuông tại A và ΔNPM vuông tại N có 

AB/NP=AC/NM

Do đó: ΔABC\(\sim\)ΔNPM

\(\widehat{A}=3\widehat{C}\)

mà \(\widehat{A}+\widehat{C}=90^0\)

nên \(\widehat{A}=67.5^0;\widehat{C}=22.5^0\)

Xét ΔABC có \(\widehat{B}>\widehat{A}>\widehat{C}\)

nên AC>BC>AB