\(\cos B=\frac{AB}{BC}=0,8\)  mà  \(\sin C=\frac{AB}{BC}=\Rightarrow\sin C=0,8\)

Theo bài ra ta có :

\(\sin C^2+\cos C^2=\frac{AB}{BC}^2+\frac{AC}{BC}^2\)

\(=\frac{\left(AB^2+AC^2\right)}{BC^2}\)

\(=\frac{BC^2}{BC^2}\)

\(=1\)

\(\Rightarrow\cos C^2=1-\sin C^2=1-0,8^2=0,36\)

\(\Rightarrow\cos C=0,6\)hoặc \(\cos C=-0,6\)( loại vì C là một góc nhọn )

\(\Rightarrow\cos C=0,6\)

\(\Rightarrow\tan C=\frac{0,8}{0,6}=\frac{4}{3};\cot C=\frac{0,6}{0,8}=0,75\)

Vậy : \(\cos C=0,6\); \(\tan C=\frac{4}{3}\)và \(\cot C=0,75\)

ta co : \(\sin^2B+\cos^2B=1\)

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

\(\Rightarrow\sin^2B=1-\left(0,8\right)^2\)

\(\Rightarrow\sin^2B=1-0,64\)

\(\Rightarrow\sin^2B=0,36\)

\(\Rightarrow\sin B=0,6\)

ta co:   \(\tan B=\frac{\sin B}{\cos B}\)hay \(\tan B=\frac{0,6}{0,8}\)

\(\Rightarrow\tan B=0,75\)

ta co :  \(\cot B=\frac{\cos B}{\sin B}\)hay \(\cot B=\frac{0,8}{0,6}\)

\(\Rightarrow\cot B=\frac{4}{3}\)

+) \(B+C=90^0\)

\(\Rightarrow\sin B=\cos C=0,6\)

\(\Rightarrow\cos B=\sin C=0,8\)

\(\Rightarrow\tan B=\cot C=0,75\)

\(\Rightarrow\cot B=\tan C=\frac{4}{3}\)

Hình bạn tự vẽ nhé !

* Ta có : AB² = AC² + BC²

             AB² = 0,9 + 1,2 = 2,1

       ==> AB ~ 1,5 (m)

sinB = AC/AB = 0,9/1,5 = 0,6

CosB= BC/AB = 1,2/1,5=0,8

tanB= AC/BC = 0,9/1,2=0,75

cotB= BC/AC=1,2/0,9=1,3

Ta có AC vg AB

       \(BC^2\) = \(AC^2\)+ \(AB^2\)

Hay \(BC^2\) = \(0,9^2\)+ \(1,2^2\)

       \(BC^2\)=  \(2,25\)

   => \(BC\) =  \(\sqrt{2,25}\) = \(1,5\)cm

      \(\sin\widehat{B}\)= \(\frac{AC}{AB}\)=\(\frac{0,9}{1,5}\)= \(0,6\)

      \(\cos\widehat{B}\)= \(\frac{BC}{AB}\)=\(\frac{1,2}{1,5}\)= \(0,8\)

     \(\tan\widehat{B}\)= \(\frac{AC}{BC}\)= \(\frac{0,9}{1,2}\)= \(0,75\)

      \(\cot\widehat{B}\)= \(\frac{BC}{AC}\)= \(\frac{1,2}{0,9}\)= \(\frac{4}{3}\)

      \(\sin\widehat{C}\)= \(\cos\widehat{B}\)= \(0,8\)

      \(\cos\widehat{C}\)= \(\sin\widehat{B}\)= \(0,6\)

     \(\tan\widehat{C}\)= \(\cot\widehat{B}\)= \(\frac{4}{3}\)

      \(\cot\widehat{C}\)= \(\tan\widehat{B}\)= \(0,75\)

Cos A sao > 1 đc bạn ơi . Bn xem lại đề ạ

Tl

Đề bài nỗi ròi

#Kirito

cos N=4/5

=>sin P=4/5

cos P=căn 1-(4/5)^2=3/5

tan P=4/5:3/5=4/3

cot P=1:4/3=3/4

a) theo đl pytago:
AB^2+AC^2=BC^2
=> AC^2=BC^2-AB^2
=>AC^2=144
=>AC=căn 144 = 12cm
Vì BC>AC>AB=>góc A > góc B > góc C

Xet tam giac ABC co goc A = 90 do (gt)

Ta co AB^2 + AC^2 = BC^2 (dinh ly Pi-ta-go)

=>AC^2 = BC^2 - AB^2 = 15^2 - 9^2 = 225 - 81 = 144

=>AC = can bac 2 cua 144 = 12

Vi BC > AC > AB => goc A > goc B > goc C

Xet tam giac ABC co:

BA = BD (gt)  (1)

goc BAE = goc BDE = 90 do (gt)  (2)

BE (canh chung)  (3)

Tu (1), (2), (3) => tam giac EBA = tam giac EBD (canh huyen-canh goc vuong)

Cau hoi tiep theo tui bo tay.com