Question Number 170624 by cortano1 last updated on 27/May/22
Commented by som(math1967) last updated on 28/May/22
Commented by som(math1967) last updated on 28/May/22
Commented by som(math1967) last updated on 28/May/22
$${I}\:{think}\:\angle{ABC}\:\:{should}\:{be}\:\theta \\ $$
Commented by cortano1 last updated on 28/May/22
$$\angle{ABD}\:=\:\theta \\ $$
Commented by som(math1967) last updated on 28/May/22
$${if}\:\angle{ABC}=\theta\:{then}\:{I}\:{can}\:{solve}\:{the}\:{problem} \\ $$
Commented by cortano1 last updated on 28/May/22
$${ok}\:{sir}.\:{let}\:\angle{ABC}=\theta \\ $$
Commented by som(math1967) last updated on 28/May/22
$${DE}={FC}={ksin}\phi \\ $$$${BE}={kcos}\phi \\ $$$${BC}={hcot}\theta \\ $$$$\therefore{EC}={DF}={hcot}\theta−{kcos}\phi \\ $$$${AF}={h}−{FC}={h}−{ksin}\phi \\ $$$$\:\frac{{DF}}{{AF}}={cot}\alpha \\ $$$$\frac{{hcot}\theta−{kcos}\phi}{{h}−{ksin}\phi}={cot}\alpha \\ $$$${hcot}\theta−{kcos}\phi={hcot}\alpha−{kcot}\alpha{sin}\phi \\ $$$${h}\left({cot}\theta−{cot}\alpha\right)={k}\left({cos}\phi−{sin}\phi{cot}\alpha\right) \\ $$$$\:\:\boldsymbol{{h}}=\frac{\boldsymbol{{k}}\left(\boldsymbol{{cos}\phi}−\boldsymbol{{sin}\phi{cot}\alpha}\right)}{\left(\boldsymbol{{cot}\theta}−\boldsymbol{{cot}\alpha}\right)} \\ $$$$ \\ $$