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