Question Number 223139 by nECxx2 last updated on 15/Jul/25
Commented by nECxx2 last updated on 15/Jul/25
$${Please}\:{help} \\ $$
Answered by gregori last updated on 15/Jul/25
$${Let}\:{RU}={a}\:,\:{UT}\:=\:{b}\: \\ $$$$\:\Rightarrow\mathrm{25}\left({a}+{b}\right)\:=\:\mathrm{25}{a}\:+\:\mathrm{49}{b}−{ab}\left({a}+{b}\right) \\ $$$$\Rightarrow\mathrm{25}{a}+\mathrm{25}{b}\:=\:\mathrm{25}{a}+\mathrm{49}{b}−{ab}\left({a}+{b}\right) \\ $$$$\:\Rightarrow{ab}\left({a}+{b}\right)\:=\:\mathrm{24}{b}\: \\ $$$$\:\Rightarrow{a}\left({a}+{b}\right)=\:\mathrm{24}\: \\ $$
Commented by nECxx2 last updated on 16/Jul/25
$${I}\:{really}\:{dont}\:{understand}\:{your}\:{approach}. \\ $$$${please}\:{explain}\:{the}\:{concept}\:{used}. \\ $$
Commented by gregori last updated on 16/Jul/25
$${Stewart}\:{theorem} \\ $$
Answered by mr W last updated on 16/Jul/25
$${say}\:{RU}={x},\:{UT}={y} \\ $$$$\mathrm{5}^{\mathrm{2}} {x}+\mathrm{7}^{\mathrm{2}} {y}=\left({x}+{y}\right)\left(\mathrm{5}^{\mathrm{2}} +{xy}\right) \\ $$$$\Rightarrow{x}\left({x}+{y}\right)=\mathrm{24} \\ $$
Commented by mr W last updated on 16/Jul/25
Commented by mr W last updated on 16/Jul/25
$${you}\:{can}\:{get}\:{this}\:{using}\:{law}\:{of}\:{cosines}: \\ $$$$\mathrm{cos}\:\theta=\frac{{m}^{\mathrm{2}} +{d}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{md}} \\ $$$$\mathrm{cos}\:\theta'=\frac{{n}^{\mathrm{2}} +{d}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{nd}}=−\mathrm{cos}\:\theta \\ $$$$\frac{{n}^{\mathrm{2}} +{d}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{nd}}=−\frac{{m}^{\mathrm{2}} +{d}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{md}} \\ $$$$\Rightarrow{mb}^{\mathrm{2}} +{nc}^{\mathrm{2}} =\left({m}+{n}\right)\left({d}^{\mathrm{2}} +{mn}\right) \\ $$
Commented by nECxx2 last updated on 17/Jul/25
$${Thank}\:{you},\:{Mr}.\:{W} \\ $$