In the given triangle, we have
[tex]\tan(\theta) = \dfrac{20}{15} = \dfrac43[/tex]
Recall that
[tex]\sin^2(\theta) + \cos^2(\theta) = 1 \implies \tan^2(\theta) + 1 = \sec^2(\theta)[/tex]
Solving for [tex]\sec(\theta)[/tex], we have
[tex]\sec(\theta) = \sqrt{\tan^2(\theta) + 1} = \sqrt{\left(\dfrac43\right)^2 + 1} = \sqrt{\dfrac{25}9} = \boxed{\dfrac53}[/tex]
(D)
