Answer:
AC = [tex]\sqrt{15}[/tex]
Step-by-step explanation:
Assuming you require AC
Using Pythagoras' identity in the right triangle
The square on the hypotenuse is equal to the sum of the squares on the other 2 sides , that is
AC² + BC²= AB²
AC² + 7² = 8²
AC² + 49 = 64 ( subtract 49 from both sides )
AC² = 15 ( take square root of both sides )
AC = [tex]\sqrt{15}[/tex] ≈ 3.87 ( to 2 dec. places )
