Solution: Given that \(\displaystyle\triangle{A}{B}{C}\) and \(\displaystyle\triangle{D}{E}{F}\) are right triangle and \(\displaystyle\triangle{A}{B}{C}\approx\triangle{D}{E}{F}\)

Now as we know

\(\displaystyle{\cos{\theta}}={\frac{{{B}{a}{s}{e}}}{{{H}{y}{p}{o}{tan{{c}}}{o}{u}{s}}}}\)

\(\displaystyle{\cos{{B}}}={\frac{{{B}{C}}}{{{A}{B}}}}\)

Step 2

Now by similarity ratio property we know

\(\displaystyle{\frac{{{A}{B}}}{{{D}{E}}}}={\frac{{{B}{C}}}{{{E}{F}}}}\)

\(\displaystyle\Rightarrow{\frac{{{E}{F}}}{{{D}{E}}}}={\frac{{{B}{C}}}{{{A}{B}}}}\)

from (1) and (2)

\(\displaystyle{\cos{{B}}}={\frac{{{E}{F}}}{{{D}{E}}}}\)