Answer:
[tex]\sec A =\frac{\sqrt{13}}{3}[/tex]
Step-by-step explanation:
Given
[tex]\tan A = 2/3[/tex]
Required
[tex]\sec\ A[/tex]
First, we have:
[tex]\tan A = \frac{x}{y}[/tex]
Where
[tex]x \to oppo site\\[/tex]
[tex]y \to adja cent[/tex]
[tex]z \to hypotenuse[/tex]
So:
[tex]\tan A = \frac{x}{y} =\frac{2}{3}[/tex]
By comparison:
[tex]x = 2; y =3[/tex]
Using Pythagoras, we have:
[tex]z^2 = x^2 +y^2[/tex]
[tex]z^2 = 2^2 +3^2[/tex]
[tex]z^2 = 13[/tex]
[tex]z = \sqrt{13[/tex]
[tex]\sec A =\frac{z}{y}[/tex]
[tex]\sec A =\frac{\sqrt{13}}{3}[/tex]
