Complete Question:
In ∆ABC, if sin A = 4/5 and tan A = 4/3, then what is cos A?
Answer:
[tex]cos A= \frac{3}{5}[/tex]
Explanation:
Given
[tex]sin A = 4/5[/tex]
[tex]tan A = 4/3[/tex]
Required
[tex]cos A[/tex]
In trigonometry;
[tex]tanA = \frac{sinA}{cosA}[/tex]
Multiply both sides by cosA
[tex]cos A * tanA = \frac{sinA}{cosA} * cos A[/tex]
[tex]cos A * tanA = sinA[/tex]
Divide both sides by tanA
[tex]\frac{cos A * tanA}{tanA} = \frac{sinA}{tanA}[/tex]
[tex]cos A= \frac{sinA}{tanA}[/tex]
Substitute values for sinA and tanA
[tex]cos A= \frac{4/5}{4/3}[/tex]
[tex]cos A= \frac{4}{5} / \frac{4}{3}[/tex]
[tex]cos A= \frac{4}{5} * \frac{3}{4}[/tex]
[tex]cos A= \frac{4 * 3}{5 * 4}[/tex]
[tex]cos A= \frac{3}{5}[/tex]
