IN the diagram, |LN| = 4cm, LNM = 90o and tan y = 2/3. What is the area of the ∆LMN?
The correct answer is B. 12cm2
\(tan \hspace{1mm}y = \frac{opp}{adj}=\frac{2}{3}\Rightarrow \frac{2}{3}=\frac{
LN}{MN}=\frac{4}{MN}\\
â´ MN = \frac{12}{2}=6\\
Area \hspace{1mm}of \hspace{1mm}\Delta LMN = \frac{1}{2}\times MN\times LN=\frac{1}{2}\times 6\times 4=12cm\)
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