Since tan y = \(\frac{2}{3}\) and LN = 4

tan y = \(\frac{2 \times 4}{3 \times 4} = \frac{8}{12} = \frac{4}{6}\)

then tan y = \(\frac{opp}{adj}\)

MN = 6cm

Area of angle LMN = \(\frac{1}{2}\)bh

= \(\frac{1}{2} \times 6 \times 4\)

= 12cm³