Question Number 181151 by mr W last updated on 22/Nov/22
Answered by SEKRET last updated on 22/Nov/22
$$\:\sqrt{\mathrm{2}} \\ $$
Answered by SEKRET last updated on 22/Nov/22
$$ \\ $$
Commented by Rasheed.Sindhi last updated on 22/Nov/22
$${Advanced}\:{booking}\:{for}\:{answer}? \\ $$
Commented by Acem last updated on 22/Nov/22
$$\:@{Rasheed},\:{You}\:{made}\:{me}\:{laugh} \\ $$
Commented by Rasheed.Sindhi last updated on 23/Nov/22
Good morning Acem sir!
Answered by Acem last updated on 22/Nov/22
Commented by mr W last updated on 22/Nov/22
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Commented by Acem last updated on 22/Nov/22
$$\:;\:{m}=\:\ell\:\mathrm{cos}\:\beta \\ $$
Answered by mr W last updated on 22/Nov/22
Commented by Acem last updated on 22/Nov/22
$${I}\:{liked}\:{this}\:{question}! \\ $$
Commented by mr W last updated on 22/Nov/22
$$\mathrm{3}^{\mathrm{2}} +\left(\mathrm{3}+{y}\right)^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} \\ $$$$\Rightarrow{y}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\left(\mathrm{5}−\mathrm{1}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{2}} \\ $$