Question Number 146776 by mathdanisur last updated on 15/Jul/21
Commented by mr W last updated on 15/Jul/21
$${AB}^{\mathrm{2}} =\mathrm{20}^{\mathrm{2}} −{x}^{\mathrm{2}} =\left(\mathrm{20}+\mathrm{5}\right)^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} −\mathrm{2}×\mathrm{25}×\mathrm{10}\:\mathrm{cos}\:{C} \\ $$$${AD}^{\mathrm{2}} ={x}^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} −\mathrm{2}×\mathrm{5}×\mathrm{10}\:\mathrm{cos}\:{C} \\ $$$$\Rightarrow\mathrm{20}^{\mathrm{2}} −{x}^{\mathrm{2}} =\left(\mathrm{20}+\mathrm{5}\right)^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} +\mathrm{5}\left({x}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\mathrm{50} \\ $$$$\Rightarrow{x}=\mathrm{5}\sqrt{\mathrm{2}} \\ $$
Commented by mathdanisur last updated on 15/Jul/21
$${thank}\:{you}\:{Ser} \\ $$