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In-MNO-MN-6units-MO-4-units-and-NO-12-units-If-the-bisector-of-the-angle-M-meets-NO-at-P-calculate-NP-




Question Number 198054 by necx122 last updated on 09/Oct/23
In △MNO, MN=6units, MO=4 units  and NO=12 units. If the bisector of the  angle M meets NO at P, calculate NP.
$${In}\:\bigtriangleup{MNO},\:{MN}=\mathrm{6}{units},\:{MO}=\mathrm{4}\:{units} \\ $$$${and}\:{NO}=\mathrm{12}\:{units}.\:{If}\:{the}\:{bisector}\:{of}\:{the} \\ $$$${angle}\:{M}\:{meets}\:{NO}\:{at}\:{P},\:{calculate}\:{NP}. \\ $$
Commented by som(math1967) last updated on 09/Oct/23
But MN+MO=10  NO=12  MN+MO<NO  Triangle possible?
$${But}\:{MN}+{MO}=\mathrm{10} \\ $$$${NO}=\mathrm{12} \\ $$$${MN}+{MO}<{NO} \\ $$$${Triangle}\:{possible}? \\ $$
Commented by necx122 last updated on 09/Oct/23
The amswer you gave initially was the  same that the textbook gave but I didnt  understand it. Please can you or anyone  solve with a clear explanation. Thanks  in advamce.
$${The}\:{amswer}\:{you}\:{gave}\:{initially}\:{was}\:{the} \\ $$$${same}\:{that}\:{the}\:{textbook}\:{gave}\:{but}\:{I}\:{didnt} \\ $$$${understand}\:{it}.\:{Please}\:{can}\:{you}\:{or}\:{anyone} \\ $$$${solve}\:{with}\:{a}\:{clear}\:{explanation}.\:{Thanks} \\ $$$${in}\:{advamce}. \\ $$
Commented by som(math1967) last updated on 10/Oct/23
In △MNO , bisector of ∠M  cut NO at P  then ((MN)/(MO))=((NP)/(PO))  let MN=m,  MO=n  ∴ ((NP)/(PO))=(m/n)⇒NP : PO=m : n  if NO=x units  then NP=x×(m/((m+n))) units  PO=x×(n/((m+n))) units
$${In}\:\bigtriangleup{MNO}\:,\:{bisector}\:{of}\:\angle{M} \\ $$$${cut}\:{NO}\:{at}\:{P} \\ $$$${then}\:\frac{{MN}}{{MO}}=\frac{{NP}}{{PO}} \\ $$$${let}\:{MN}={m},\:\:{MO}={n} \\ $$$$\therefore\:\frac{{NP}}{{PO}}=\frac{{m}}{{n}}\Rightarrow{NP}\::\:{PO}={m}\::\:{n} \\ $$$${if}\:{NO}={x}\:{units} \\ $$$${then}\:{NP}={x}×\frac{{m}}{\left({m}+{n}\right)}\:{units} \\ $$$${PO}={x}×\frac{{n}}{\left({m}+{n}\right)}\:{units} \\ $$

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