\(\sin^2\widehat{A}+\cos^2\widehat{A}=1\Leftrightarrow\cos^2\widehat{A}=1-\left(\dfrac{3}{5}\right)^2=1-\dfrac{9}{25}=\dfrac{16}{25}\\ \Leftrightarrow\cos\widehat{A}=\dfrac{4}{5}\\ \tan\widehat{A}=\dfrac{\sin\widehat{A}}{\cos\widehat{A}}=\dfrac{3}{4}\\ \Rightarrow\cot\widehat{A}=\dfrac{1}{\tan\widehat{A}}=\dfrac{4}{3}\)

Bài 2: 

\(\cos\widehat{A}=\dfrac{3\sqrt{39}}{20}\)

\(\tan\widehat{A}=\dfrac{7}{20}:\dfrac{3\sqrt{39}}{20}=\dfrac{7}{3\sqrt{39}}=\dfrac{7\sqrt{39}}{117}\)

\(\cot\widehat{A}=\dfrac{3\sqrt{39}}{7}\)

tana = 3/4.
=>cota=1/ tana =1:3/4=4/3
sina /cosa =tana
=> sina =tana .cosa =3/4. cosa
lại có sin^2(a)+cos^2(a)=1
<=>9/16cos^2(a)+cos^2=1
<=>25/16cos^2(a)=1
<=>cos^2(a)=16/25
=>[cosa =4/5=>sina =3/5
    [cosa =-4/5=> sina =-2/5

\(\cos^2a+sin^2a=1 ;\frac{\sin a}{\cos a}=\tan a\)

Thanks

ΔABC vuông tại B, áp dụng hệ thức lượng trong tam giác vuông ta có :

\(\frac{1}{BH^2}=\frac{1}{AB^2}+\frac{1}{BC^2}\Rightarrow BH=\sqrt{\frac{\left(AB\cdot BC\right)^2}{AB^2+BC^2}}=7,2\left(cm\right)\)

a) Xét ΔHBA có ^AHB = 90⁰ ( BH ⊥ AC ) => ΔHBA vuông tại H

Khi đó ta có : \(\sin A=\frac{BH}{AB}=\frac{4}{5};\cos A=\frac{AH}{AB}=\frac{\sqrt{AB^2-BH^2}}{AB}=\frac{3}{5};\tan A=\frac{BH}{AH}=\frac{BH}{\sqrt{AB^2-BH^2}}=\frac{4}{3}\)

(bonus cho cotgA) : \(\cot A=\frac{AH}{BH}=\frac{\sqrt{AB^2-BH^2}}{BH}=\frac{3}{4}\)

b) Vì ΔHBA vuông tại H (cmt) => ^A + ^ABH = 90⁰

Khi đó : \(\sin\widehat{ABH}=\cos A=\frac{3}{5};\cos\widehat{ABH}=\sin A=\frac{4}{5};\tan\widehat{ABH}=\cot A=\frac{3}{4};\cot\widehat{ABH}=\tan A=\frac{4}{3}\)

\(\sin^2\widehat{A}+\cos^2\widehat{A}=1\Leftrightarrow\cos^2\widehat{A}=1-\dfrac{16}{25}=\dfrac{9}{25}\\ \Leftrightarrow\cos\widehat{A}=\dfrac{3}{5}\\ \tan\widehat{A}=\dfrac{\sin\widehat{A}}{\cos\widehat{A}}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\\ \cot\widehat{A}=\dfrac{1}{\tan\widehat{A}}=\dfrac{3}{4}\)

\(\sin A=0,8\Rightarrow A=arcsin0,8_{ }\)

\(\Rightarrow\cos A=cos\left(arcsin0,8\right)=\dfrac{3}{5}\)

     tanA=tan(arcsin0,8)=4/3

     cotA=1:4/3=3/4