Question

1 In XYY, XYZ= 90 and
xy + yz = 289 Find xz
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Answer 1

Answer:

According to me it s

${xy}^{2} + {yz}^{2} = 289$

not xy and yz

if so the answer is:-

As /_XYZ=90°

:. By Pythagoras theorem

${ xz}^{2} = {xy}^{2} + {yz}^{2} \\ {xz}^{2} = 289 \\ xz = \sqrt{289} \\ xz = 17$

:. xz =17