Question Number 103672 by M±th+et+s last updated on 16/Jul/20
Commented by M±th+et+s last updated on 16/Jul/20
$${if} \\ $$$${AD}+{AC}={BC} \\ $$$${and} \\ $$$${AB}+{BD}={AC} \\ $$$$ \\ $$$${find}\:\angle\:{ABC} \\ $$$$ \\ $$
Commented by som(math1967) last updated on 16/Jul/20
$$\mathrm{If}\:\mathrm{ABC}\:\mathrm{is}\:\mathrm{triangle}\:\mathrm{then}\:\mathrm{how} \\ $$$$\mathrm{AB}+\mathrm{BC}=\mathrm{AC}\:? \\ $$
Commented by M±th+et+s last updated on 16/Jul/20
$${sorry}\:{sir}\:{its}\:{typo}\: \\ $$$${thank}\:{you}\:{for}\:{your}\:{comment} \\ $$
Answered by Worm_Tail last updated on 16/Jul/20
$$\:\:{AC}^{\mathrm{2}} ={BC}^{\mathrm{2}} +{BA}^{\mathrm{2}} −\mathrm{2}\left({BC}\right)\left({BA}\right){cos}\left({A}\hat {{B}C}\right)\:\:{cosine}\:\:{rule} \\ $$$$\:\:\left({AB}+{BC}\right)^{\mathrm{2}} ={BC}^{\mathrm{2}} +{BA}^{\mathrm{2}} −\mathrm{2}\left({BC}\right)\left({BA}\right){cos}\left({A}\hat {{B}C}\right)\:\:{given} \\ $$$$\:\:{AB}^{\mathrm{2}} +\mathrm{2}\left({AB}\right)\left({BC}\right)+{BC}^{\mathrm{2}} ={BC}^{\mathrm{2}} +{BA}^{\mathrm{2}} −\mathrm{2}\left({BC}\right)\left({BA}\right){cos}\left({A}\hat {{B}C}\right)\: \\ $$$$\:\:\mathrm{2}\left({AB}\right)\left({BC}\right)=−\mathrm{2}\left({BC}\right)\left({BA}\right){cos}\left({A}\hat {{B}C}\right)\: \\ $$$$\:\:−\mathrm{1}={cos}\left({A}\hat {{B}C}\right)\:\Rightarrow{A}\hat {{B}C}=\mathrm{180}° \\ $$$$ \\ $$$$ \\ $$