Answer:
MO = 8 units
MP = 4 units
Step-by-step explanation:
[tex] In\: \triangle MOP, \\
\angle P = 90\degree ... (given) \\
\angle M = 60\degree ... (given) \\
\therefore \angle O = 30\degree(3^{rd} \: \angle \: of\: \triangle) \\
Let \huge\purple {MO = x \: units}... (1)\\
\therefore MP = \frac{1}{2} \times MO\\(side\: opposite \: to\: 30\degree) \\\\
\therefore MP = \frac{1}{2} \times x\\\\
\huge\red {\therefore MP = \frac{1}{2} x}.... (2)\\\\
\therefore PO = \frac{\sqrt 3}{2} \times MO\\(side\: opposite \: to\: 60\degree) \\\\
\therefore PO = \frac{\sqrt 3}{2} \times x\\\\
\huge\orange{\therefore PO = \frac{\sqrt 3}{2}x} \\\\
\because MP + PO + MO = P(\triangle MOP) \\\\
\therefore \frac{1}{2} x+\frac{\sqrt 3}{2}x+ x = 12+4\sqrt 3\\\\
\therefore \frac{1}{2} x+x+\frac{\sqrt 3}{2}x= 4(3+\sqrt 3)\\\\
\therefore \frac{3}{2} x+\frac{\sqrt 3}{2}x= 4(3+\sqrt 3)\\\\
\therefore \frac{(3+\sqrt 3)}{2} x= 4(3+\sqrt 3)\\\\
\therefore x = 4(3+\sqrt 3)\times \frac{2}{(3+\sqrt 3)}\\\\
\therefore x = 4\cancel{(3+\sqrt 3)}\times \frac{2}{\cancel {(3+\sqrt 3)}}\\\\
\therefore x = 4\times 2\\\\
\therefore x = 8\\\\
\huge\purple {\boxed{\implies MO = 8}} \: \\ [From\: equation \: (1)]\\\\
\because MP = \frac{1}{2} x \:
\\ [From\: equation \: (2)]\\\\
\therefore MP = \frac{1}{2} \times 8\\\\
\huge\red {\boxed{\therefore MP = 4 \: units}}
[/tex]
