In triangle ABC, if [tex]\sin \angle A>0[/tex] and [tex]\tan\angle A>0,[/tex] then [tex]\cos \angle A>0.[/tex]
Use trigonometric formula
[tex]\tan \alpha =\dfrac{\sin \alpha}{\cos \alpha}[/tex]
to find [tex]\cos \angle A:[/tex]
[tex]\cos \angle A=\dfrac{\sin \alpha}{\tan \alpha}=\dfrac{\frac{4}{5}}{\frac{4}{3}}=\dfrac{3}{5}.[/tex]
Answer: [tex]\cos \angle A=\dfrac{3}{5}.[/tex]
