Answer:
BC = [tex]\frac{12}{13}[/tex]
Step-by-step explanation:
Using Pythagoras' identity in the right triangle
BC² + AC² = AB²
BC² + ([tex]\frac{5}{13}[/tex] )² = 1²
BC² + [tex]\frac{25}{169}[/tex] = 1 ( subtract [tex]\frac{25}{169}[/tex] from both sides )
BC² = [tex]\frac{169}{169}[/tex] - [tex]\frac{25}{169}[/tex] = [tex]\frac{144}{169}[/tex] ( take square root of both sides )
BC = [tex]\sqrt{\frac{144}{169} }[/tex] = [tex]\frac{12}{13}[/tex]
