Answer:
The answer is
[tex]x = 12 \\ y = 20 \\ z = 15[/tex]
Step-by-step explanation:
By using Pythagoras low
Then we can find that
[tex] {y}^{2} = {x}^{2} + {16}^{2} \\ {z}^{2} = {x}^{2} + {9}^{2} \\ {y}^{2} + {z}^{2} = {25}^{2} \\ then \\ {x}^{2} + {9}^{2} + {x}^{2} + {16}^{2} = {25}^{2} \\ 2 {x}^{2} = 288 \\ x = 12 \\ By \: substituting \: x \: into \: the \: previous \: equations \\ then \: \\ y = 20 \\ z = 15[/tex]
