Áp dụng hệ thức lượng trong $ΔMNP$ vuông tại $M$ ta được:
$MK^{2} = NK.KP$
⇒ $MK = \sqrt{NK.KP} = \sqrt{4.9} = \sqrt{36} = 6 cm$
$MN^{2} = NK.NP
⇒ $MN = \sqrt{NK.NP} = \sqrt{4.(4+9)} = \sqrt{52} = 2\sqrt{13} cm$
Ta có:
\(\left\{ \begin{array}{l}sin\widehat{N}=\dfrac{MK}{MN}=\dfrac{6}{2\sqrt{13}}=\dfrac{3\sqrt{13}}{13}\\cos\widehat{N}=\dfrac{NK}{MN} = \dfrac{4}{2\sqrt{13}} = \dfrac{2\sqrt{13}}{13}\\tan\widehat{N} = \dfrac{sin\widehat{N}}{cos\widehat{N}} = \dfrac{\dfrac{3\sqrt{13}}{13}}{\dfrac{2\sqrt{13}}{13}}=\dfrac{3}{2}\\cot\widehat{N} = \dfrac{1}{tan\widehat{N}}= \dfrac{1}{\dfrac{3}{2}}= \dfrac{2}{3}\end{array} \right.\)
