Solution:
Given the right triangle;
The sum of angles in a triangle is 180 degrees. Thus;
[tex]\begin{gathered} z^o+60^o+90^o=180^o \\ \\ z^o=180^o-60^o-90^o \\ \\ z=30^o \end{gathered}[/tex]
Using cosine trigonometry ratio;
[tex]\cos\theta=\frac{adjacent}{hypotenuse}[/tex]
Given;
[tex]\theta=60^o,adjacent=\sqrt{3},hypotenuse=m[/tex]
Thus;
[tex]\begin{gathered} \cos60^o=\frac{\sqrt{3}}{m} \\ \\ m=\frac{\sqrt{3}}{\cos60^o} \\ \\ m=2\sqrt{3} \end{gathered}[/tex]
Lastly, using Pythagorean Theorem;
[tex](opposite)^2=(hypotenuse)^2-(adjacent)^2[/tex][tex]\begin{gathered} p^2=m^2-(\sqrt{3})^2 \\ \\ p^2=(2\sqrt{3})^2-(\sqrt{3})^2 \\ \\ p^2=12-3 \\ \\ p^2=9 \\ \\ p=\sqrt{9} \\ \\ p=\pm3 \\ \\ p=3 \end{gathered}[/tex]
ANSWERS:
[tex]\begin{gathered} z=30^o \\ \\ m=2\sqrt{3} \\ \\ p=3 \end{gathered}[/tex]
