Question

Anyone can find “v” and “u”??

Answer 1

Answer:

$v = \frac{16}{3} \: \: and \: \: u = \frac{20}{3}$

Step-by-step 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.

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

$= > {v}^{2} + 6v + 9 = 41 + {v}^{2}$

$= > {v}^{2} - {v}^{2} + 6v = 41 - 9$

$= > 6v = 32$

$= > v = \frac{32}{6} = \frac{16}{3}$

Putting the value of V in eqn. below :-

${u}^{2} = 16 + { (\frac{16}{3} )}^{2}$

$= > {u}^{2} = \frac{256}{9} + 16$

$= > {u}^{2} = \frac{256 + 144}{9} = \frac{400}{9}$

$= > u = \sqrt{ \frac{400}{9} } = \frac{20}{3}$