Question

Solve for z,m and p. Type answer as whole number.

Answer 1

Given the right triangle;

The sum of angles in a triangle is 180 degrees. Thus;

`\begin{gathered} z^o+60^o+90^o=180^o \\ \\ z^o=180^o-60^o-90^o \\ \\ z=30^o \end{gathered}`

Using cosine trigonometry ratio;

`\cos\theta=(adjacent)/(hypotenuse)`

Given;

`\theta=60^o,adjacent=√(3),hypotenuse=m`

Thus;

`\begin{gathered} \cos60^o=(√(3))/(m) \\ \\ m=(√(3))/(\cos60^o) \\ \\ m=2√(3) \end{gathered}`

Lastly, using Pythagorean Theorem;

`(opposite)^2=(hypotenuse)^2-(adjacent)^2`
`\begin{gathered} p^2=m^2-(√(3))^2 \\ \\ p^2=(2√(3))^2-(√(3))^2 \\ \\ p^2=12-3 \\ \\ p^2=9 \\ \\ p=√(9) \\ \\ p=\pm3 \\ \\ p=3 \end{gathered}`

ANSWERS:

`\begin{gathered} z=30^o \\ \\ m=2√(3) \\ \\ p=3 \end{gathered}`