Question

In the diagram, given below, triangle ABC is right-angled at B and BD is perpendicular to AC. Find: $\cos \angle DBC$

Answer 1

$\begin{array}{c}\cos \angle DBC=\frac{DB}{BC}\\ =\frac{\left(\frac{60}{13}\right)}{5}\\ =\frac{12}{13}\end{array}$