Question

Anyone can find “v” and “u”??

Answer 1

`v = (16)/(3) \: \: and \: \: u = (20)/(3)`

Explanation:

In triangle WYZ ,

`{u}^(2) = {4}^(2) + {v}^(2) = 16 + {v}^(2)`

In traingle WXZ ,

`{(3 + v)}^(2) = {5}^(2) + {u}^(2)`

Putting the value of U^2 in above eqn.

`= > {(3 + v)}^(2) = 25 + (16 + {v}^(2) )`

`= > {v}^(2) + 6v + 9 = 41 + {v}^(2)`

`= > {v}^(2) - {v}^(2) + 6v = 41 - 9`

`= > 6v = 32`

`= > v = (32)/(6) = (16)/(3)`

Putting the value of V in eqn. below :-

`{u}^(2) = 16 + { ((16)/(3) )}^(2)`

`= > {u}^(2) = (256)/(9) + 16`

`= > {u}^(2) = (256 + 144)/(9) = (400)/(9)`

`= > u = \sqrt{ (400)/(9) } = (20)/(3)`