Question Number 83369 by Power last updated on 01/Mar/20
Commented by mr W last updated on 01/Mar/20
$$\angle{A}=\mathrm{90}° \\ $$$$\Rightarrow{AD}={BD}={DC}=\mathrm{5} \\ $$
Commented by mathmax by abdo last updated on 01/Mar/20
$${we}\:{use}\:{the}\:{formulae}\:{AB}^{\mathrm{2}} \:+{AC}^{\mathrm{2}} =\mathrm{2}{AD}^{\mathrm{2}} \:+\frac{{BC}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{8}^{\mathrm{2}} \:+\mathrm{6}^{\mathrm{2}} =\mathrm{2}{AD}^{\mathrm{2}} \:+\frac{\mathrm{100}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{2}{AD}^{\mathrm{2}} =\mathrm{100}−\mathrm{50}\:=\mathrm{50}\:\Rightarrow{AD}^{\mathrm{2}} =\mathrm{25}\:\Rightarrow \\ $$$${AD}=\mathrm{5} \\ $$
Answered by mind is power last updated on 01/Mar/20
$${cos}\left({B}\right)=\frac{\mathrm{10}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} }{\mathrm{2}.\mathrm{10}.\mathrm{8}}\:=\frac{\mathrm{128}}{\mathrm{160}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${AD}^{\mathrm{2}} =\mathrm{8}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{2}.\mathrm{8}.\mathrm{5}{cos}\left({B}\right)=\mathrm{64}+\mathrm{25}−\mathrm{64}=\mathrm{25} \\ $$$${AD}=\mathrm{5} \\ $$