a: Xét ΔABC vuông tại A có \(\hat{B}+\hat{C}=90^0\)

=>cot C=tan B=2

Ta có: \(\tan C\cdot\cot C=1\)

=>\(\tan C=\frac12\)

Ta có: \(1+\tan^2C=\frac{1}{cos^2C}\)

=>\(\frac{1}{cos^2C}=1+\left(\frac12\right)^2=\frac54\)

=>\(cos^2C=\frac45\)

=>\(cosC=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}\)

Ta có: \(\sin^2C+cos^2C=1\)

=>\(\sin^2C=1-\frac45=\frac15\)

=>sin C=\(\frac{1}{\sqrt5}\)

b: \(\sin^2B+cos^2B=1\)

=>\(cos^2B=1-\left(\frac{\sqrt3}{2}\right)^2=1-\frac34=\frac14\)

=>\(cosB=\frac12\)

tan B=\(\frac{\sin B}{cosB}=\frac{\sqrt3}{2}:\frac12=\sqrt3\)

\(cotB=\frac{1}{\tan B}=\frac{1}{\sqrt3}\)

a: Xét ΔABC vuông tại A có \(\hat{B}+\hat{C}=90^0\)

=>cot C=tan B=2

Ta có: \(\tan C\cdot\cot C=1\)

=>\(\tan C=\frac12\)

Ta có: \(1+\tan^2C=\frac{1}{cos^2C}\)

=>\(\frac{1}{cos^2C}=1+\left(\frac12\right)^2=\frac54\)

=>\(cos^2C=\frac45\)

=>\(cosC=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}\)

Ta có: \(\sin^2C+cos^2C=1\)

=>\(\sin^2C=1-\frac45=\frac15\)

=>sin C=\(\frac{1}{\sqrt5}\)

b: \(\sin^2B+cos^2B=1\)

=>\(cos^2B=1-\left(\frac{\sqrt3}{2}\right)^2=1-\frac34=\frac14\)

=>\(cosB=\frac12\)

tan B=\(\frac{\sin B}{cosB}=\frac{\sqrt3}{2}:\frac12=\sqrt3\)

\(cotB=\frac{1}{\tan B}=\frac{1}{\sqrt3}\)

\(a,\sin\widehat{C}=\dfrac{AB}{BC};\cos\widehat{C}=\dfrac{AC}{BC};\tan\widehat{C}=\dfrac{AB}{AC};\cot\widehat{C}=\dfrac{AC}{AB}\\ b,BC=\sqrt{AB^2+AC^2}=13\left(cm\right)\left(pytago\right)\\ \Rightarrow\sin\widehat{B}=\dfrac{AC}{BC}=\dfrac{12}{13};\cos\widehat{B}=\dfrac{AB}{BC}=\dfrac{5}{13}\\ \tan\widehat{B}=\dfrac{AC}{AB}=\dfrac{12}{5};\cot\widehat{B}=\dfrac{AB}{AC}=\dfrac{5}{12}\)

\(\tan\widehat{B}=\dfrac{AC}{AB}=\dfrac{12}{5}\approx\tan67^022'\\ \Rightarrow\widehat{B}\approx67^022'\\ \Rightarrow\widehat{C}=90^0-67^022'=22^038'\)

Đề sai rồi bạn

ΔABC vuông tại A

=>sin B=cos C

=>\(cosC=0,6\)

Ta có: \(\sin^2C+cos^2C=1\)

=>\(\sin^2C=1-0,6^2=1-0,36=0,64=0,8^2\)

=>sin C=0,8

tan C=sin C:cosC=0,8:0,6=4/3

cot C=cosC/sinC=0,6/0,8=3/4

ΔABC vuông tại A

=>sin B=cos C

=>\(cosC=0,6\)

Ta có: \(\sin^2C+cos^2C=1\)

=>\(\sin^2C=1-0,6^2=1-0,36=0,64=0,8^2\)

=>sin C=0,8

tan C=sin C:cosC=0,8:0,6=4/3

cot C=cosC/sinC=0,6/0,8=3/4

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

\(\sin\widehat{B}=\cos\widehat{C}=\dfrac{AC}{BC}=\dfrac{4}{5}\\ \cos\widehat{B}=\sin\widehat{C}=\dfrac{AB}{BC}=\dfrac{3}{5}\\ \tan\widehat{B}=\cot\widehat{C}=\dfrac{AC}{AB}=\dfrac{4}{3}\\ \cot\widehat{B}=\tan\widehat{C}=\dfrac{AB}{AC}=\dfrac{3}{4}\)

Câu a) với b) tính cos, tan, sin là tính góc hay cạnh vậy cậu?

a) Ta có: \(cos\alpha=\dfrac{12}{13}\)

Mà: \(sin^2\alpha+cos^2a=1\)

\(\Rightarrow sin^2\alpha=1-cos^2\alpha\)

\(\Rightarrow sin^2\alpha=1-\left(\dfrac{12}{13}\right)^2\)

\(\Rightarrow sin^2\alpha=\dfrac{25}{169}\)

\(\Rightarrow sin\alpha=\sqrt{\dfrac{25}{169}}\)

\(\Rightarrow sin\alpha=\dfrac{5}{13}\)

Mà: \(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{5}{13}}{\dfrac{12}{13}}=\dfrac{5}{12}\)

b) Ta có: \(cos\alpha=\dfrac{3}{5}\)

Mà: \(sin^2\alpha+cos^2\alpha=1\)

\(\Rightarrow sin^2\alpha=1-cos^2\alpha\)

\(\Rightarrow sin^2\alpha=1-\left(\dfrac{3}{5}\right)^2\)

\(\Rightarrow sin^2\alpha=\dfrac{16}{25}\)

\(\Rightarrow sin\alpha=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}\)

Mà: \(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{4}{5}}{\dfrac{3}{5}}=\dfrac{4}{3}\)

2:

a: BC=căn 16^2+12^2=20cm

Xét ΔABC vuông tại A có

sin B=cos C=AC/BC=3/5

cos B=sin C=AB/BC=4/5

tan B=cot C=3/5:4/5=3/4

cot B=tan C=1:3/4=4/3

b: AH=căn 13^2-5^2=12cm

Xét ΔAHC vuông tại H có

sin C=AH/AC=12/13

=>cos B=12/13

cos C=HC/AC=5/13

=>sin B=5/13

tan C=12/13:5/13=12/5

=>cot B=12/5

tan B=cot C=1:12/5=5/12

c: BC=3+4=7cm

AB=căn BH*BC=2*căn 7(cm)

AC=căn CH*BC=căn 21(cm)

Xét ΔABC vuông tại A có

sin B=cos C=AC/BC=căn 21/7

sin C=cos B=AB/BC=2/căn 7

tan B=cot C=căn 21/7:2/căn 7=1/2*căn 21

cot B=tan C=1/căn 21/2=2/căn 21