Since we are dealing with a right triangle, we can use the following trigonometric identities
[tex]\sin \theta=\frac{O}{H},\cos \theta=\frac{A}{H}[/tex]
Where θ is an inner angle (different than 90°) of the triangle, O is the opposite side to θ, A is the adjacent side to θ, and H is the hypotenuse.
a) In our case,
[tex]\begin{gathered} \theta=30\text{degre}e \\ H=14,A=m,O=n \\ \Rightarrow\sin (30degree)=\frac{n}{14} \\ \Rightarrow n=14\cdot\sin (30degree)=14\cdot0.5=7 \\ \Rightarrow n=7 \end{gathered}[/tex]
and
[tex]\begin{gathered} \Rightarrow\cos (30degree)=\frac{m}{14} \\ \Rightarrow m=14(\cos (30degree))=14\cdot\frac{\sqrt[]{3}}{2}=7\sqrt[]{3} \\ \Rightarrow m=7\sqrt[]{3} \end{gathered}[/tex]
The answers are n=7 and m=7sqrt(3).
3) In a diagram, the problem states
Using the same trigonometric identities mentioned in part 1) (plus the tangent function), we get
[tex]\begin{gathered} \sin (30degree)=\frac{18}{H},\tan (30degree)=\frac{18}{A} \\ \Rightarrow H=\frac{18}{\sin(30degree)},A=\frac{18}{\tan(30degree)}=\frac{18}{\frac{1}{\sqrt[]{3}}}=18\sqrt[]{3} \\ \Rightarrow H=\frac{18}{0.5}=36,A=18\sqrt[]{3} \\ \Rightarrow H=36,A=18\sqrt[]{3} \end{gathered}[/tex]
The hypotenuse is equal to 36 ft, and the other leg is equal to 18sqrt(3) ft
