here to this question we have to use Pythagoras theorem that is
[tex] {a}^{2} + {b}^{2} = {hypotnuse}^{2} [/tex]
here,
a=6.7
b=x (cuz' we don't know it's value)
c= hypotenuse =16
now, substitute these values in the above formula then we get,
[tex] {6.7}^{2} + {x}^{2} = {16}^{2} \\ = 14.529[/tex]
as the answer should be given to 1 decimal place we get 14.529
