Đáp án:
i) $NF = 15;\widehat F \approx {36^0}52';\widehat N = {53^0}8'$
ii) $MO = \dfrac{{36}}{5}; FO=\dfrac{{48}}{5}$
Giải thích các bước giải:
i) Ta có:
$\begin{array}{l}
\Delta MNF;\widehat {NMF} = {90^0};MO \bot NF = O;MN = 9cm;MF = 12cm\\
\Rightarrow \left\{ \begin{array}{l}
NF = \sqrt {M{N^2} + M{F^2}} = \sqrt {{9^2} + {{12}^2}} = 15\\
\sin F = \dfrac{{MN}}{{NF}} = \dfrac{3}{5} \Rightarrow \widehat F = \arcsin \left( {\dfrac{3}{5}} \right) \Rightarrow \widehat F \approx {36^0}52'\\
\widehat N = {90^0} - \widehat F = {53^0}8'
\end{array} \right.
\end{array}$
ii) $\begin{array}{l}
\Delta MNF;\widehat {NMF} = {90^0};MO \bot NF = O;MN = 9cm;MF = 12cm\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{1}{{M{O^2}}} = \dfrac{1}{{M{F^2}}} + \dfrac{1}{{M{N^2}}} = \dfrac{1}{{{{12}^2}}} + \dfrac{1}{{{9^2}}} \Rightarrow MO = \dfrac{{36}}{5}\\
FO = \sqrt {M{F^2} - M{O^2}} = \sqrt {{{12}^2} - {{\left( {\dfrac{{36}}{5}} \right)}^2}} = \dfrac{{48}}{5}
\end{array} \right.
\end{array}$
