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Question-146776




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

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