Solution
Finding Z:
[tex]\begin{gathered} \text{ The sum of angles in a triangle is 180\degree} \\ \text{ Thus, we have:} \\ \\ 60\degree+90\degree+\angle Z\degree=180\degree \\ 150\degree+\angle Z=180\degree \\ \text{ Subtract 150 from both sides} \\ \\ \therefore\angle Z=180-150=30\degree \end{gathered}[/tex]
Finding P:
[tex]\begin{gathered} \text{ Applying SOHCAHTOA, we have that:} \\ \tan60\degree=\frac{P}{\sqrt{3}} \\ \\ \therefore P=\sqrt{3}\times\tan60\degree \\ \\ P=\sqrt{3}\times\sqrt{3} \\ \\ P=3 \end{gathered}[/tex]
Finding M:
[tex]\begin{gathered} \text{ Applying SOHCAHTOA once more, we have:} \\ \sin Z=\frac{\sqrt{3}}{m} \\ \\ \text{ We know Z= 30} \\ \\ \sin30\degree=\frac{\sqrt{3}}{m} \\ \\ \text{ We can rewrite this as:} \\ m=\frac{\sqrt{3}}{\sin30\degree} \\ \\ \text{ But,} \\ \sin30\degree=\frac{1}{2} \\ \\ \text{ Thus,} \\ m=\frac{\sqrt{3}}{\frac{1}{2}}=2\sqrt{3} \\ \\ m=2\sqrt{3} \end{gathered}[/tex]
